The given question is incomplete, the complete question is:
A chemistry student weighs out 0.0349g of formic acid HCHO2 into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1500M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.
Answer:
The correct answer is 5.06 ml.
Explanation:
Based on the given information, the weight of formic acid given is 0.0349 grams. The volume of formic acid of V1 given is 250 ml. The molecular mass of formic acid is 46 grams per mole. Now the molarity of formic acid will be,
[HCOOH] = weight * 1000 / molecular mass * volume (ml)
= 0.0349 * 1000 / 46 * 250
= 0.003035 M or M1
The molarity of NaOH given is 0.1500 M or M2
Let us assume that the volume needed to attain equivalence point is V2 ml. The volume V2 can be determined by using the dilution equation,
M1V1 = M2V2
V2 = M1V1/M2
V2 = 0.003035 * 250 / 0.1500
V2 = 5.06 ml.
Hence, the volume of NaOH needed is 5.06 ml.
A 1.0 kg object absorbs 1,303 J of heat energy and experiences a temperature increase of 5.2∘C. What is the object’s specific heat, in joules per gram-degree celsius? Report your answer with the correct number of significant figures.
Answer:
c = 250.58 J/kg/[tex]^{0}C[/tex]
Explanation:
The specific heat of a substance is the required quantity of heat to increase or decrease the temperature of its unit mas by 1 kelvin.
Q = mcΔθ
where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is its specific heat and Δθ is the change in temperature of the substance.
Given that; m = 1.0 kg, Q = 1303 J and Δθ = 5.2 [tex]^{0}C[/tex], then;
c = Q ÷ (mΔθ)
= 1303 ÷ (1.0 × 5.2)
= 1303 ÷ 5.2
= 250.58 J/kg/[tex]^{0}C[/tex]
The specific heat of the object is 250.58 J/kg/[tex]^{0}C[/tex].
Answer:
0.25
Explanation:
The specific rotation of (S)-carvone (at 20°C) is +61. A chemist prepared a mixture of (R)-carvone and its enantiomer, and this mixture had an observed rotation of -55°.
A) What is the specific rotation of (R)-carvone at 20°C?
B) Calculate the % ee of this mixture.
C) What percentage of the mixture is (S)-carvone?
Answer:
a) Specific Rotation of (R)-carvone = -61°
b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%
c) The percentage of (S)-carvone in the mixture = 4.9%
Explanation:
a) The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.
Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.
Specific Rotation of (R)-carvone = -(61°) = -61°
b) Enantiometic excess is used to measure the optical purity of an enantiomeric mixture.
The enantiomeric excess is given mathematically as
ee% = (Observed rotation × 100)/(Specific rotation)
Hence, to calculate the enantiomeric excess of (R)-carvone,
Observed rotation of the mixture = -55°
Specific Rotation of (R)-carvone = -61°
ee% = (-55×100)/(-61) = 90.16% = 90.2%
c) An enantiomeric excess of 90.2% for (R)-carvone indicates that it's actual percentage is 90.2% more than the percentage of its enantiomeric partner, (S)-carvone, in the mixture.
Let the percentage of (R)-carvone in the mixture be x
Let the percentage of (S)-carvone in the mixture be y
x + y = 100
x - y = 90.2
2x = 190.2
x = (190.2/2) = 95.1%
y = 100 - x = 100 - 95.1 = 4.9%
Hence, the percentage of (R)-carvone in the mixture = 95.1%
The percentage of (S)-carvone in the mixture = 4.9%
Hope this Helps!!!
a) Specific Rotation of (R)-carvone = -61°
b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%
c) The percentage of (S)-carvone in the mixture = 4.9%
a) Calculation of Specific Rotation:The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.
Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.
Specific Rotation of (R)-carvone = -(61°) = -61°
b) Calculation for Enantiomeric excess:
The enantiomeric excess is given mathematically as
ee% = (Observed rotation × 100)/(Specific rotation)
Hence, to calculate the enantiomeric excess of (R)-carvone,
Observed rotation of the mixture = -55°
Specific Rotation of (R)-carvone = -61°
ee% = (-55×100)/(-61) = 90.16% = 90.2%
c) Calculation of percentage:
Let the percentage of (R)-carvone in the mixture be x
Let the percentage of (S)-carvone in the mixture be y
x + y = 100
x - y = 90.2
2x = 190.2
x = (190.2/2) = 95.1%
y = 100 - x = 100 - 95.1 = 4.9%
Hence, the percentage of (R)-carvone in the mixture = 95.1%
The percentage of (S)-carvone in the mixture = 4.9%
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Limiting reagent problem. How many grams of H2O is produced from 40.0 g N2O4 and 25.0 g N2H4. N2O4 (l) + 2 N2H4 (l) → 3 N2 (g) + 4 H2O(g)
Answer:
28.13 g of H2O.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N2O4(l) + 2N2H4 (l) → 3N2(g) + 4H2O(g)
Next, we shall determine the masses of N2O4 and N2H4 that reacted and the mass of H2O produced from the balanced equation.
This is illustrated below:
Molar mass of N2O4 = (14x2) + (16x4) = 92 g/mol
Mass of N2O4 from the balanced equation = 1 x 92 = 92g
Molar mass of N2H4 = (14x2) + (4x1) = 32 g/mol
Mass of N2H4 from the balanced equation = 2 x 32 = 64 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 4 x 18 = 72 g
Summary:
From the balanced equation above,
92 g of N2O4 reacted with 64 g of N2H4 to produce 72 g of H2O.
Next, we shall determine the limiting reactant.
This can be obtained as follow:
From the balanced equation above,
92 g of N2O4 reacted with 64 g of N2H4.
Therefore, 40 g of N2O4 will react with = (40 x 64)/92 = 27.83 g of N2H4.
From the calculations made above, we can see that it will take a higher mass i.e 27.83 g than what was given i.e 25 g of N2H4 to react completely with 40 g of N2O4.
Therefore, N2H4 is the limiting reactant and N2O4 is the excess reactant.
Finally, we shall determine the mass of H2O produced from the reaction of 40.0 g of N2O4 and 25.0 g of N2H4.
In this case the limiting reactant will be used because it will produce the maximum amount of H2O as all of it is consumed in the reaction.
The limiting reactant is N2H4 and the mass of H2O produced can be obtained as follow:
From the balanced equation above,
64 g of N2H4 reacted to produce 72 g of H2O.
Therefore, 25 g of N2H4 will react to produce = (25 x 72)/64 = 28.13 g of H2O.
Therefore, 28.13 g of H2O were obtained from the reaction.
A student mixes wants to prepare 24.1 mmol of benzamide from benzoyl chloride and NH4OH. If the student uses excess 15 M NH4OH, how many mL of Benzoyl chloride must be used
Answer:
2.81mL
Explanation:
Based on the reaction:
C₆H₃COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl
Benzoyl chloride + ammonia → Benzamide
1 mole of benzoyl chloride in excess of ammonia produce 1 mole of Benzamide.
Thus, assuming a theoretical yield, to produce 24.1mmoles of benzamide you require 24.1mmoles of benzoyl chloride.
As molar mass of benzoyl chloride is 141g/mol, mg you require are:
mg Benzoyl chloride: 24.1mmol × (141mg / 1mmol) = 3398.1mg = 3.3981g of benzoyl chloride.
to convert this mass to mL, you require density of Benzoyl chloride (1.21g/mL). Thus, mL you need are:
3.3981g × (1mL / 1.21g) =
2.81mLAn aqueous solution containing 5.06 g of lead(II) nitrate is added to an aqueous solution containing 6.03 g of potassium chloride.Enter the balanced chemical equation for this reaction. Be sure to include all physical states.balanced chemical equation:What is the limiting reactant?lead(II) nitratepotassium chlorideThe percent yield for the reaction is 82.9% . How many grams of the precipitate are formed?precipitate formed:gHow many grams of the excess reactant remain?excess reactant remaining:
Answer:
Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)
3.52 g of PbCl2
3.76 g of KCl
Explanation:
The equation of the reaction is;
Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)
Number of moles of Pb(NO3)2 =mass/molar mass 5.06g/331.2 g/mol = 0.0153 moles
Number of moles of KCl= mass/ molar mass= 6.03g/74.5513 g/mol= 0.081 moles
Next we obtain the limiting reactant; the limiting reactant yields the least number of moles of products.
For Pb(NO3)2;
1 mole of Pb(NO3)2 yields 1 mole of PbCl2
Therefore 0.0153 moles of Pb(NO3)2 yields 0.0153 moles of PbCl2
For KCl;
2 moles of KCl yields 1 mole of PbCl2
0.081 moles of KCl yields 0.081 moles ×1/2 = 0.041 moles of PbCl2
Therefore Pb(NO3)2 is the limiting reactant.
Theoretical Mass of precipitate obtained = 0.0153 moles of PbCl2 × 278.1 g/mol = 4.25 g of PbCl2
% yield = actual yield/theoretical yield ×100
Actual yield = % yield × theoretical yield /100
Actual yield= 82.9 ×4.25/100
Actual yield = 3.52 g of PbCl2
If 1 mole of Pb(NO3) reacts with 2 moles of KCl
0.0153 moles of Pb(NO3)2 reacts with 0.0153 moles × 2 = 0.0306 moles of KCl
Amount of excess KCl= 0.081 moles - 0.0306 moles = 0.0504 moles of KCl
Mass of excess KCl = 0.0504 moles of KCl × 74.5513 g/mol = 3.76 g of KCl
The ionization constant of lactic acid ch3ch(oh) co2h am acid found in the blood after strenuous exercise is 1.36×10^-4 If 20.0g of latic acid is used to make a solution with a volume of 1.00l what is the concentration of hydronium ion in the solution
Answer:
Explanation:
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
ionisation constant = 1.36 x 10⁻⁴ .
molecular weight of lactic acid = 90 g
moles of acid used = 20 / 90
= .2222
it is dissolved in one litre so molar concentration of lactic acid formed
C = .2222M
Let n be the fraction of moles ionised
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionisation constant Ka
Ka = nC x nC / C - nC
= n²C ( neglecting n in the denominator )
n² x .2222 = 1.36 x 10⁻⁴
n = 2.47 x 10⁻²
nC = 2.47 x 10⁻² x .2222
= 5.5 x 10⁻³
So concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per litre .
The concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per liter .
Ionization of lactic acid can be represented as:
CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻ + H⁺
Given:
ionization constant = 1.36 x 10⁻⁴
mass= 20.0 g
Now, Molecular weight of lactic acid = 90 g
[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]
It is dissolved in 1.00L so molar concentration of lactic acid formed will be
C = 0.22M
Consider "n" to be the fraction of moles ionized
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionization constant Ka
[tex]K_a =\frac{nC*nC}{C-nC}[/tex]
[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )
On substituting the values we will get:
[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]
To find the concentration of hydronium ion in the solution,
[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]
So, concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per liter.
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below are three reactions showing how chlorine from CFCs (chlorofluorocarbons) destroy ozone (O3) in the stratosphere. Ozone blocks harmful ultraviolet radiation from reaching earth’s surface. Show how these 3 equations sum to produce the net equation for the decomposition of two moles of ozone to make three moles of diatomic oxygen (2 O3→ 3 O2), and calculate the enthalpy change. (6 points) R1 O2 (g) → 2 O (g) ΔH1°= 449.2 kJ R2 O3 (g) + Cl (g) → O2 (g) + ClO (g) ΔH2° = -126 kJ R3 ClO (g) + O (g) → O2 (g) + Cl (g) ΔH3°= -268 kJ
Answer:
ΔH = -338.8kJ
Explanation:
it is possible to sum the enthalpy changes of some reactions to obtain the enthalpy change of the whole reaction (Hess's law).
Using the reactions:
R₁ O₂(g) → 2O(g) ΔH₁°= 449.2 kJ
R₂ O₃(g) + Cl(g) → O₂(g) + ClO(g) ΔH₂° = -126 kJ
R₃ ClO (g) + O (g) → O₂ (g) + Cl (g) ΔH₃°= -268 kJ
By the sum 2R₂ + 2R₃:
(2R₂ + 2R₃) = 2O(g) + 2O₃(g) → 4O₂(g)
ΔH = 2ₓ(-126kJ) + (2ₓ-268kJ) = -788kJ
Now, this reaction + R₁
2O₃(g) → 3O₂(g)
ΔH = -768kJ + 449.2kJ
ΔH = -338.8kJWhen the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:
P2O5 (s) + H2O (l) =H3PO4 (aq)
The balanced chemical equation for the reaction between hydrogen sulfide and oxygen is:
2H2S(g) + 3O2(g) =2H2O(l) + 2SO2(g)
We can interpret this to mean:
3moles of oxygen and_______moles of hydrogen sulfide react to produce______moles of water and_______ moles of sulfur dioxide.
Answer:
1. The coefficients are: 1, 3, 2
2. From the balanced equation, we obtained the following:
3 moles oxygen, O2 reacted.
2 moles of Hydrogen sulfide, H2S reacted.
2 moles of water were produced.
2 moles of sulphur dioxide, SO2 were produced.
Explanation:
1. Determination of the coefficients of the equation.
This is illustrated below:
P2O5(s) + H2O(l) <==> H3PO4(aq)
There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:
P2O5(s) + H2O(l) <==> 2H3PO4(aq)
There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:
P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)
Now the equation is balanced.
The coefficients are: 1, 3, 2.
2. We'll begin by writing the balanced equation for the reaction. This is given below:
2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)
From the balanced equation above,
3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.
A compound X has a molecular ion peak in its mass spectrum at m/z 136. What information does this tell us about X
Explanation:
The mass to charge ratio =136
Carbon-14 has a half-life of 5720 years and this is a fast-order reaction. If a piece of wood has converted 75 % of the carbon-14, then how old is it?
Answer:
11445.8years
Explanation:
Half-life of carbon-14 = 5720 years
First we have to calculate the rate constant, we use the formula :
Arrange the following oxides in order of increasing acidity.
Rank from least acidic to most acidic. To rank items as equivalent,overlap them.
CaO
P2O5
SO3
SiO2
Al2O3
CO2
Answer:
Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.
With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:
CaO > Al2O3 > SiO2 > CO2 > P2O5 > SO3. Hence, CaO is the least acidic and SO3 is the most acidic.
Since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:
[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]
The least acidic is CaOThe most acidic is [tex]SO_3[/tex]Note the following:
Acidity of an oxide depends on its electronegativity.Non-metals are more electronegative, while metals are less electronegative.Acidity of oxides increases across a period as you move from left to the right side of a periodic table.Acidity of oxides decreases down a group (column) in a periodic table.Using the periodic table diagram given in the attachment below, we can rank the given oxides according to their increasing acidity.
CaO, is the least, because it is an oxide of the metal, Calcium, which is at the far left in group 2 in the periodic table.The next is, [tex]Al_2O_3[/tex]. Aluminum is a metal from group 3.[tex]SiO_2[/tex] is an oxide of Silicon, also in group 4 but below Carbon.[tex]CO_2[/tex] is an oxide of Carbon, from group 4.
[tex]P_2O_5[/tex] is an oxide of the non-metal, Phosphorus, a group 5 element
[tex]SO_3[/tex] is an oxide of the non-metal, Sulphur, a group 6 element.
Therefore, since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:
[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]
The least acidic is CaOThe most acidic is [tex]SO_3[/tex]Learn more here:
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What class of organic product results when 1-heptyne is treated with a mixture of mercuric acetate (HgSO4) in aqueous sulfuric acid (H2O/H2SO4)
Answer:
heptan-2-one
Explanation:
In this case, the final product would be a ketone: heptan-2-one. To understand why this molecule is produced we have to check the reaction mechanism.
The first step is the protonation of the triple bond to produce the more stable carbocation (a secondary one) by the action of sulfuric acid [tex]H_2SO_4[/tex]. The next step is the attack of water to the carbocation to produce a new bond between C and the O, producing a positive charge in the oxygen. Then, a deprotonation step takes place to produce an enol. Finally, we will have a rearrangement (keto-enol tautomerism) to produce the final ketone.
See figure 1
I hope it helps!
can I get some urgent help please?
Answer:
hi here goes your answer
Explanation:
iv. The lower the PH, the weaker the base
What is the molar mass of a protein if a solution of 0.020 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 ∘ C
Answer:
26.5 kD
Explanation:
Here we can apply the formula ∏ = iMRT, where ∏ = osmotic pressure = 0.56 - ( given ). This is only one part of the information we are given / can conclude in this case ....
i = van’t Hoff factor = 1 for a protein molecule,
R = gas constant = 62.36 L torr / K-mol,
T ( temperature in Kelvin ) = 25 + 273 - conversion factor C° + 273 = 298K
( Known initially ) ∏ = osmotic pressure = 0.56 torr
..... besides the part " M " in the formula, which we have no information on whatsoever, as we have to determine it's value.
_____
Substitute derived / known values to solve for M ( moles / liter ) -
∏ = iMRT
⇒ 0.56 = ( 1 )( M )( 62.36 )( 298 )
⇒ 0.56 = M( 18583.28 )
⇒ M = 0.56 / 18583.28 ≈ 0.00003013461 ....
_____
We know that M = moles / liter, so we can use this to solve for moles, and hence calculate the molar mass by the formula molar mass = g / mol -
M = mol / l
⇒ 0.00003013461 = 0.020 / 25 mL ( 0.025 L ),
0.020 / 0.025 = 0.8 g / L
⇒ 0.8 g = 0.00003013461 moles,
molar mass = 0.8 g / 0.00003013461 moles = 26,548 g / mol = 26.5 kD
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast furnace. After cooling the pure liquid iron, Steve determines that he has produced 288g of iron ingots. What is the theoretical yield of liquid iron, in grams? Just enter a numerical value. Do not enter units.
Answer: 313.6
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe_2O_3=\frac{450g}{160g/mol}=2.8moles[/tex]
[tex]\text{Moles of} CO=\frac{260g}{28g/mol}=9.3moles[/tex]
[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)[/tex]
According to stoichiometry :
1 mole of [tex]Fe_2O_3[/tex] require 3 moles of [tex]CO[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] will require=[tex]\frac{3}{1}\times 2.8=8.4moles[/tex] of [tex]CO[/tex]
Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]CO[/tex] is the excess reagent.
As 1 mole of [tex]Fe_2O_3[/tex] give = 2 moles of [tex]Fe[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] give =[tex]\frac{2}{1}\times 2.8=5.6moles[/tex] of [tex]Fe[/tex]
Mass of [tex]Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g[/tex]
Theoretical yield of liquid iron is 313.6 g
4 Al + 3O2 → 2Al2O3 If 14.6 grams Al are reacted, how many liters of O2 at STP would be required?
Answer: 9.08 L
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles[/tex]
[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]
According to stoichiometry :
4 moles of [tex]Al[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 0.54 moles of [tex]Al[/tex] will require=[tex]\frac{3}{4}\times 0.54=0.405moles[/tex] of [tex]O_2[/tex]
Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = 1 atm
V= Volume of the gas = ?
T= Temperature of the gas = 273 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= 0.405
[tex]V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L[/tex]
Thus 9.08 L of [tex]O_2[/tex] at STP would be required
Considering the reaction stoichiometry and STP conditions, 9.072 L of O₂ at STP would be required.
The balanced reaction is:
4 Al + 3 O₂ → 2 Al₂O₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
Al: 4 moles O₂: 3 moles Al₂O₃: 2 moles
Being 27 g/mole the molar mass of Al, this is the amount of mass that a substance contains in one mole, then if 14.6 grams Al are reacted, the number of moles of Al that react is calculated as:
[tex]14.6 gramsx\frac{1 mole}{27 grams}= 0.54 moles[/tex]
Then you can apply the following rule of three: if by stoichiometry 4 moles of Al react with 3 moles of O₂, 0.54 moles of Al react with how many moles of O₂?
[tex]amount of moles of O_{2} =\frac{0.54 moles of Alx3 moles of O_{2} }{4 moles of Al}[/tex]
amount of moles of O₂= 0.405 moles
On the other side, the STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
Then you can apply the following rule of three: if by definition of STP 1 mole of O₂ occupies 22.4 L, 0.405 moles of O₂, how much volume does it occupy?
[tex]volume=\frac{0.405 moles of O_{2}x22.4 L }{1 mole of O_{2} }[/tex]
volume= 9.072 L
Finally, 9.072 L of O₂ at STP would be required.
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brainly.com/question/16487206?referrer=searchResults brainly.com/question/14446695?referrer=searchResults brainly.com/question/11564309?referrer=searchResults brainly.com/question/4025026?referrer=searchResults brainly.com/question/18650135?referrer=searchResultscetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce of water.
Answer:
0.60 mol
Explanation:
There is some info missing. I think this is the original question.
Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of water.
Step 1: Given data
Moles of water required: 1.5 mol
Step 2: Write the balanced equation
C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)
Step 3: Calculate the moles of oxygen needed to produce 1.5 mol of water
The molar ratio of O₂ to H₂O is 2.5:1. The moles of oxygen needed to produce 1.5 mol of water are (1/2.5) × 1.5 mol = 0.60 mol
If enough experimental data supports a hypothesis, then it
Answer:
Then the hypothesis is proved and becomes a theory.
If not, then another hypothesis should be proposed and tested.
The table below shows the electronegativity values of various elements on the periodic table. Electronegativities A partial periodic table. Which pair of atoms would form a covalent bond ? calcium (Ca) and bromine (Br) rubidium (Rb) and sulfur (S) cesium (Cs) and nitrogen (N) oxygen (O) and chlorine (Cl)
Answer:
Oxygen and Chlorine
Explanation:
Covalent bonds involve the sharing of electrons between nonmetals.
Answer:
oxygen (O) and chlorine (Cl)
Explanation:
cuz i said so
How many mL of calcium hydroxide are required to neutralize 25.0 mL of 0.50 M
nitric acid?
Answer:
6.5 mL
Explanation:
Step 1: Write the balanced reaction
Ca(OH)₂ + 2 HNO₃ ⇒ Ca(NO₃)₂ + 2 H₂O
Step 2: Calculate the reacting moles of nitric acid
25.0 mL of 0.50 M nitric acid react.
[tex]0.0250L \times \frac{0.50mol}{L} = 0.013 mol[/tex]
Step 3: Calculate the reacting moles of calcium hydroxide
The molar ratio of Ca(OH)₂ to HNO₃ is 1:2. The reacting moles of Ca(OH)₂ are 1/2 × 0.013 mol = 6.5 × 10⁻³ mol
Step 4: Calculate the volume of calcium hydroxide
To answer this, we need the concentration of calcium hydroxide. Since the data is missing, let's suppose it is 1.0 M.
[tex]6.5 \times 10^{-3} mol \times \frac{1,000mL}{1.0mol} = 6.5 mL[/tex]
Night vision glasses detect
energy emitted from cooling objects?
ultraviolet
infrared
X-ray
Answer:
I think the answer is " Night vision glasses detect Infrared" energy emitted from cooling objects.
Explanation:
16. A metal element and a non-metal element are brought near each other and allowed to react. What's the most likely type of compound
that will form between these two elements?
A. lonic and covalent
B. lonic
C. Covalent
D. Neither, metals and non-metals don't react.
Answer:
B) Ionic
Explanation:
Air contains nitrogen, oxygen, argon, and trace gases. Ifthe partial pressure of nitrogen is 592 mm Hg, oxygen is160 mm Hg, argon is 7 mm Hg, and trace gas is 1 mm Hg,what is the atmospheric pressure
Answer:
760 mmHg
Explanation:
Step 1: Given data
Partial pressure of nitrogen (pN₂): 592 mmHgPartial pressure of oxygen (pO₂): 160 mmHgPartial pressure of argon (pAr): 7 mmHgPartial pressure of the trace gas (pt): 1 mmHgStep 2: Calculate the atmospheric pressure
Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.
P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg
When a solution is diluted with water, the ratio of the initial to final
volumes of solution is equal to the ratio of final to initial molarities
Select one:
True
-
When a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities. The statement is True.
Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.
There are various methods of expressing the concentration of a solution.
Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.
Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.
Learn more about Concentrations, here:
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Drag each image to the correct location on the model. Each image can be used more than once. Apply the rules and principles of electron configuration to draw the orbital diagram of aluminum. Use the periodic table to help you.
Answer:
The answer to your question is given below.
Explanation:
Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:
Al (13) —› 1s² 2s²2p⁶ 3s²3p¹
The orbital diagram is shown on the attached photo.
Answer: screen shot
Explanation:
Identify a homogeneous catalyst:
a. SO2 over vanadium (V) oxide
b. H2SO4 with concentrated HCl
c. Pd in H2 gas
d. N2 and H2 catalyzed by Fe
e. Pt with methane
Answer:
b, H2SO4 with HCl, as they are both liquid acids
Suppose you titrate 25.00 mL of 0.200 M KOBr with 0.200M H2SO4. The pH at half-equivalence point is 7.75 a). What is the initial pH of the 25.00mL of 0.200M KOBr mentioned above
Answer:
Approximately [tex]10.88[/tex].
Explanation:
Equilibrium constant[tex]\rm OBr^{-}[/tex] can act as a weak Bronsted-Lowry base:
[tex]\rm OBr^{-}\; (aq) + H_2O\; (l) \rightleftharpoons HOBr\; (l) + OH^{-}\; (aq)[/tex].
(Side note: the state symbol of [tex]\rm HOBr[/tex] in this equation is [tex]\rm (l)[/tex] (meaning liquid) because [tex]\rm HOBr[/tex] is a weak acid.)
However, the equilibrium constant of this reaction, [tex]K_\text{eq}[/tex], isn't directly given. The idea is to find [tex]K_\text{eq}[/tex] using the [tex]\rm pH[/tex] value at the half-equivalence point. Keep in mind that this system is at equilibrium all the time during the titration. If temperature stays the same, then the same [tex]K_\text{eq}[/tex] value could also be used to find the [tex]\rm pH[/tex] of the solution before the acid was added.
At equilibrium:
[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].
At the half-equivalence point of this titration, exactly half of the base, [tex]\rm OBr^{-}[/tex], has been converted to its conjugate acid, [tex]\rm HOBr[/tex]. Therefore, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should both be equal to one-half the initial concentration of [tex]\rm OBr^{-}[/tex].
As a result, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should be the same. The expression for [tex]K_\text{eq}[/tex] can thus be simplified:
[tex]\begin{aligned}& K_\text{eq} \\&= \frac{\left(\text{half-equivalence $[\rm HOBr\; (l)]$}\right)\cdot \left(\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\right)}{\text{half-equivalence $[\rm OBr^{-}\; (l)]$}}\\ &=\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\end{aligned}[/tex].
In other words, the [tex]K_\text{eq}[/tex] of this system is equal to the [tex]\rm OH^{-}[/tex] concentration at the half-equivalence point. Assume that [tex]\rm p\mathnormal{K}_\text{w}[/tex] the self-ionization constant of water, is [tex]14[/tex]. The concentration of [tex]\rm OH^{-}[/tex] can be found from the [tex]\rm pH[/tex] value:
[tex]\begin{aligned}& \text{half-equivalence $[\rm OH^{-}\; (aq)]$} \\ &= 10^{\rm pH - p\mathnormal{K}_\text{w}}\;\rm mol \cdot L^{-1} \\ &= 10^{7.75 - 14}\; \rm mol \cdot L^{-1}\\ &= 10^{-6.25}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
Therefore, [tex]\begin{aligned} K_\text{eq} &= 10^{-6.25}\end{aligned}[/tex].
Initial pH of the solutionAgain, since [tex]\rm KOBr[/tex] is a soluble salt, all that [tex]0.200\; \rm M[/tex] of [tex]\rm KOBr[/tex] in this solution will be in the form of [tex]\rm K^{+}[/tex] and [tex]\rm OBr^{-}[/tex] ions. Before any hydrolysis takes place, the concentration of [tex]\rm OBr^{-}[/tex] should be equal to that of [tex]\rm KOBr[/tex]. Therefore:
[tex]\text{$[\rm OBr^{-}\; (aq)]$ before hydrolysis} = 0.200\; \rm M[/tex].
Let the equilibrium concentration of [tex][\rm OH^{-}\; (aq)][/tex] be [tex]x\; \rm M[/tex]. Create a RICE table for this reversible reaction:
[tex]\begin{array}{c|ccccccc} & \rm OBr^{-}\; (aq) &+&\rm H_2O\; (l)& \rightleftharpoons & \rm HOBr\; (l)& + & \rm OH^{-}\; (aq) \\ \textbf{I}& 0.200\; \rm M & & & & 0 \; \rm M & & 0\; \rm M \\ \textbf{C} & -x\; \rm M & & & & +x \; \rm M & & +x\; \rm M \\ \textbf{E}& (0.200 + x)\; \rm M & & & & x \; \rm M & & x\; \rm M \end{array}[/tex].
Assume that external factors (such as temperature) stays the same. The [tex]K_\text{eq}[/tex] found at the half-equivalence point should apply here, as well.
[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].
At equilibrium:
[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x}[/tex].
Assume that [tex]x[/tex] is much smaller than [tex]0.200[/tex], such that the denominator is approximately the same as [tex]0.200[/tex]:
[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x} \approx \frac{x^2}{0.200}[/tex].
That should be equal to the equilibrium constant, [tex]K_\text{eq}[/tex]. In other words:
[tex]\displaystyle \frac{x^2}{0.200} \approx K_\text{eq} = 10^{-6.25}[/tex].
Solve for [tex]x[/tex]:
[tex]x \approx 3.35\times 10^{-4}[/tex].
In other words, the [tex]\rm OH^{-}[/tex] before acid was added was approximately [tex]3.35\times 10^{-4}\; \rm M[/tex], which is the same as [tex]3.35\times 10^{-4}\; \rm mol \cdot L^{-1}[/tex]. Again, assume that [tex]\rm p\mathnormal{K}_\text{w} = 14[/tex]. Calculate the [tex]\rm pH[/tex] of that solution:
[tex]\begin{aligned}\rm pH &= \rm p\mathnormal{K}_\text{w} + \log [\mathrm{OH^{-}}] \approx 10.88\end{aligned}[/tex].
(Rounded to two decimal places.)
what is the meaning of the word tetraquark?
Answer:
A tetraquark in physics is an exotic meson composed of four valence quarks.
Explanation:
It has been suspected to be allowed by quantum chromodynamics, the modern story of strong interactions.
Hope it helps.
What do chemists use percent yield calculations for in the real world?
A. To balance the reaction equation.
B. To determine how much product they will need.
C. To determine how efficient reactions are.
D. To determine how much reactant they need.
Answer:
C. To determine how efficient reactions are.
D. To determine how much reactant they need.
Explanation:
When you are doing a reaction, you are hoping for a percent yield to close of 100%. You make the reaction and determine how many product you obtain. If you know the percent yield of a reaction you can calculate the amount of reactant you need to obtain a determined amount of product.
Having this in mind:
A. To balance the reaction equation. false. To calculate percent yield you need to balance the reaction before. You don't use percent yield to balance the reaction
B. To determine how much product they will need. false. You determine how much product you obtain after the reaction. How much product you need is independent of percent yield
C. To determine how efficient reactions are. true. A way to determine efficience of a reaction is with percent yield. An efficient reaction has a high percent yield.
D. To determine how much reactant they need. true. If you know percent yield of a reaction you can know how many reactant you must add to obtain the amount of product you want.
Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the ions that would be present in solution. Be sure to consider both the coefficients and subscripts of the molecular equation, and then write this precipitation reaction in the form of a balanced complete ionic equation. Express your answer as a chemical equation including phases.
Answer:
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
Explanation:
K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)
The complete ionic equation for the above equation can be written as follow:
In solution, K2CO3 and CuF will dissociate as follow:
K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)
CuF(aq) —› Ca^2+(aq) + 2F¯(aq)
Thus, we can write the complete ionic equation for the reaction as shown below:
K2CO3(aq) + 2CuF(aq) —›
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)