Answer:
2 moles of HNO3
Explanation:
The equation seems to be balanced correctly. The problem is we done know what you started with. We will assume it is 3 moles of NO2.
If that is the case then 2 moles of HNO3 will be produced.
When silver nitrate is added to an aqueous solution of magnesium chloride, a precipitation reaction occurs that produces silver chloride and magnesium nitrate. When enough AgNO3 is added so that 34.3 g of MgCl2 react, what mass of the AgCl precipitate should form
Answer:
103.62 g of AgCl.
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
2AgNO3 + MgCl2 —> 2AgCl + Mg(NO3)2
Step 2:
Determination of the mass of MgCl2 that reacted and the mass of AgCl produced from the balanced equation.
This is illustrated below:
Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol
Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g
Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol
Mass of AgCl from the balanced equation = 2 x 143.5 = 287 g
Thus, from the balanced equation above,
95 g of MgCl2 reacted to produce 287 g of AgCl.
Step 3:
Determination of the mass of AgCl produced from the reaction of 34.3 g of MgCl2.
The mass of AgCl produced from the reaction can be obtained as follow:
Form the balanced equation above,
95 g of MgCl2 reacted to produce 287 g of AgCl.
Therefore, 34.3 g of MgCl2 will react to produce = (34.3 x 287)/95 = 103.62 g of AgCl.
Therefore, 103.62 g of AgCl were produced from the reaction.
Which functional group does the molecule below have?
A. Ether
B. Ester
C. Hydroxyl
D. Amino
Answer:
Hydroxyl
Explanation:
A hydroxyl group is a functional group that attaches to some molecules containing an oxygen and hydrogen atom, bonded together. Also spelled hydroxy, this functional group provides important functions to both alcohols and carboxylic acids.
The functional groups are the part of the organic chemistry that confers the characteristic feature of a molecule. The molecule has a hydroxyl group in its structure. Thus, option C is correct.
What are hydroxyl functional groups?Hydroxyl functional groups are the atoms or molecules that provide a distinctive property to a compound. It has a chemical formula of -OH that has oxygen covalently bonded to the hydrogen atom.
The hydroxyl group is called the alcohol group that is seen in methanol, ethanol, propanol, etc. The presence of hydrogen allows the compound to form a water bond with other molecules and makes them soluble and polar.
Therefore, option C. the molecule has a hydroxyl or alcoholic functional group attached to its carbon atom.
Learn more about the hydroxyl functional group here:
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At what pressure would 11.1 moles of a gas occupy 44.8 L at 300 K?
Answer:
[tex]P=6.10atm[/tex]
Explanation:
Hello,
In this case, we can study the ideal gas equation that relates temperature, volume, pressure and moles as shown below:
[tex]PV=nRT[/tex]
Thus, since we are asked to compute the pressure y simply solve for it as follows:
[tex]P=\frac{nRT}{V}=\frac{11.1mol*0.082\frac{atm*L}{mol*K}*300K}{44.8L}\\ \\P=6.10atm[/tex]
Best regards.
In each of the three reactions between NaOH and HCl, the sign of q for the water was positive. This means the the sign of q for the reaction was ______ and the reaction was ______.
Answer:
This means the the sign of q for the reaction was _NEGATIVE _____ and the reaction was _EXOTHERMIC_____.
Explanation:
In calorimetry, when heat is absorbed by the solution, the q-value of the solution will have a positive value. This means that the reaction will produce heat for the solution to absorb and thus the q-value for the reaction will be negative. This is an exothermic reaction.
Whereas, when heat is absorbed from the solution, the q-value for the solution will have a negative value. This means that the reaction will absorb heat from the solution and so the reaction is endothermic, and q value for the reaction is positive.
So, from the question, since the q-value of water is positive, it means that heat is absorbed by the solution and the reaction will produce a negative value of q and it's an exothermic reaction because the reaction produces heat for the solution.
A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 32.3 mg produced 87.7 mg of CO2 and 18.0 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.
Answer:
Empirical formula: C₅H₅O
Molecular formula: C₁₀H₁₀O₂
Explanation:
When a compound containing C, H and O elements is combusted, the general reaction is:
CₐHₓOₙ + O₂ → a CO₂ + X/2 H₂O
Thus, you can find moles of carbon and hydrogen knowing moles of CO₂ and H₂O that are produced.
Moles CO₂ = Moles C = 0.0877g × (1mol / 44g) =
2.0x10⁻³ moles of CO₂ = moles C
Moles H₂O = 1/2 Moles H = 0.018g × (1mol / 18g) =
1x10⁻³ moles of H₂O; 2.0x10⁻³ moles H
The mass of the moles of C and H are:
2x10⁻³ moles C ₓ (12g / mol) = 0.024g C
2x10⁻³ moles H ₓ (1g / mol) = 0.002g H
Thus, mass of Oxygen is 32.3mg - 24mg C - 2mg O = 6.3mg O
Moles are:
0.0063g O ₓ (1mol / 16g) = 4x10⁻⁴ moles O
Empirical formula is the simplest ratio of atoms in a compound. Dividing each amount of moles for each atom in the 4x10⁻⁴ moles of oxygen (The lower moles), you will obtain:
C: 2.0x10⁻³ / 4x10⁻⁴ = 5
H: 2.0x10⁻³ / 4x10⁻⁴ = 5
O: 4x10⁻⁴ / 4x10⁻⁴ = 1
Thus, empirical formula is:
C₅H₅OThe molar mass of the empirical formula is:
12×5 + 1×5 + 16×1 = 81g/mol
As molar mass of the compound is 162g/mol, molecular formula is twice empirical formula:
C₁₀H₁₀O₂Consider the following reaction: Br2(g) + 3 F2(g) LaTeX: \rightarrow→ 2 BrF3(g) LaTeX: \Delta H_{rxn}Δ H r x n= ‒836 kJ/mol Bond Bond Energy (kJ/mol) Br–Br 193 F–F 155 Using the above bond dissociation energies, calculate the energy, in kJ/mol, of a Br–F bond.
Answer: The energy of a Br–F bond is 110 kJ/mol
Explanation:
The balanced chemical reaction is,
[tex]Br_2(g)+3F_2(g)\rightarrow 2BrF_3(g)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)][/tex]
[tex]\Delta H=[(n_{Br_2}\times B.E_{Br_2})+(n_{F_2}\times B.E_{F_2}) ]-[(n_{BrF_3}\times B.E_{BrF_3})][/tex]
[tex]\Delta H=[(n_{Br_2}\times B.E_{Br-Br})+(n_{F_2}\times B.E_{F_F}) ]-[(n_{BrF_3}\times 3\times B.E_{Br-F})][/tex]
where,
n = number of moles
Now put all the given values in this expression, we get
[tex]\Delta H=[(1\times 193)+(3\times 155)]-[(2\times 3\times B.E_{Br-F})][/tex]
[tex]B.E_{Br-F}=110kJ/mol[/tex]
Thus the energy, in kJ/mol, of a Br–F bond is 110
Rank the compounds in each set in order of increasing acid strength.
(a) CH3CH2COOH CH3CHBrCOOH CH3CBr2COOH
(b) CH3CH2CH2CHBrCOOH CH3CH2CHBrCH2COOH CH3CHBrCH2CH2COOH
Answer:
See explanation
Explanation:
For this question, we have to remember the effect of an atom with high electronegativity as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an inductive effect. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be weaker and the compound will be more acid (because is easier to produce the hydronium ion [tex]H^+[/tex]).
With this in mind, for A in the last compound, we have 2 Br atoms near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have more acidity. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.
In B, the difference between the molecules is the position of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a higher inductive effect and more acidity.
See figure 1
I hope it helps!
A student mixes 2.83 mL of benzoyl chloride with excess 15 M NH4OH to produce 1.95 g of benzamide. What is the percent yield of this student's experiment
Answer:
Explanation:
The reaction of benzoyl chloride with NH₄OH to produce benzamide is:
Benzoyl chloride + ammonia → Benzamide + NH₄Cl
Molar mass of benzoyl chloride: 140.57 g/mol. Density 1.21g/mL
Molar mass benzamide: 121.14g/mol.
To know percent yield you must know the theoretical yield of the reaction (How many grams are produced assuming a yield of 100%). Percent yield will be (Actual yield / Theoretical Yield) ₓ 100
Moles of 2.83mL of benzoyl chloride are:
2.83mL ₓ (1.21g/mL) ₓ (1mol / 140.57g) = 0.02436 moles of benzoyl chloride.
As 1 mole of benzoyl chloride produce 1 mole of benzamide (Theoretical yield), theoretical moles of benzamide produced are 0.02436. In mass:
0.02436 moles ₓ (121.14g / mol) = 2.95g of benzoyl chloride
As there are produced just 1.95, percent yield is:
(1.95g / 2.95g) ₓ 100 = 66.1%
Content attribution
QUESTION 2 • 1 POINT
Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?
The given question is incomplete. The complete question is :
Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?
a) [tex]O^{2-}[/tex]
b) [tex]F^{-}[/tex]
c) [tex]N^{3-}[/tex]
d) [tex]S^{2-}[/tex]
Answer: b) [tex]F^{-}[/tex]
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here potassium is having an oxidation state of +1 called as cation and thus is an anion must have an oxidation state of -1 if they have to combine in 1: 1 ratio to give neutral ionic compound.
Thus the anion has to be [tex]F^-[/tex] which combines with [tex]K^+[/tex] in 1: 1 ratio to give [tex]KF[/tex]
Calculate the maximum volume in mL of 0.18 M HCl that a tablet containing 340 mg Al(OH)3 and 516 mg Mg(OH)2 would be expected to neutralize. Assume complete neutralization.
Answer:
171 mL of HCl
Explanation:
The first thing we want to do is consider the reaction between Al(OH)3 and water - as that is the expected reaction that is taking place,
Al(OH)3 + 3HCl → AlCl3 + 3H2O
Knowing this, let's identify the mass of Al(OH)3. Aluminum = 27 g / mol, Oxygen( 3 ) = 16 [tex]*[/tex] 3 = 48, Hydrogen ( 3 ) = 1 [tex]*[/tex] 3 = 3 - 27 + 48 + 3 = 78 g / mol. This value is approximated however ( 78 g / mol ), as the molar mass of each substance is rounded as well. Another key thing we need to do here is to convert 340 mg → grams, considering that that unit is a necessity with respect to moles, as you might know - 340 mg = 0.340 g.
Now we can calculate how much moles of HCl will be present in solution, provided we have sufficient information for that,
(0.340 g Al(OH)3) / (78.0036 g / mol Al(OH)3) [tex]*[/tex] (3 mol HCl / 1 mol Al(OH)3)
⇒ (.004358773185 g^2 / mol Al(OH)3) [tex]*[/tex] (3 HCl / Al(OH)3 )
⇒ .01307632 mol HCl
We can apply this same concept on the reaction of Mg(OH)2 and water, receiving the number of moles of HCl when that takes place. Then we can add the two ( moles of HCl ) and divide by the value " 0.18 mol / L " given to us.
" Mg(OH)2 + 2HCl → MgCl2 + 2H2O "
Molar mass of Mg(OH)2 = 58.3197 g / mol,
516 mg = 0.516 g
(0.516 g Mg(OH)2) / (58.3197 g / mol Mg(OH)2) [tex]*[/tex] (2 mol HCl / 1 mol Mg(OH)2)
= .017695564 mol HCL
___________
( .01307632 + .017695564 ) / ( 0.18 M HCl )
= 0.170954911 L
= 171 mL of HCl
Modern atomic theory states that atoms are neutral. How is this neutrality achieved in atoms? (2 points)
A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What is the molarity of this solution? Express your answer to four significant figures and include the appropriate units.
Answer:
Approximately [tex]1.854\; \rm mol\cdot L^{-1}[/tex].
Explanation:
Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.
Formula mass of strontium hydroxideLook up the relative atomic mass of [tex]\rm Sr[/tex], [tex]\rm O[/tex], and [tex]\rm H[/tex] on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.
[tex]\rm Sr[/tex]: [tex]87.62[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].[tex]\rm H[/tex]: [tex]1.008[/tex].Calculate the formula mass of [tex]\rm Sr(OH)_2[/tex]:
[tex]M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}[/tex].
Number of moles of strontium hydroxide in the solution[tex]M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}[/tex] means that each mole of [tex]\rm Sr(OH)_2[/tex] formula units have a mass of [tex]121.634\; \rm g[/tex].
The question states that there are [tex]10.60\; \rm g[/tex] of [tex]\rm Sr(OH)_2[/tex] in this solution.
How many moles of [tex]\rm Sr(OH)_2[/tex] formula units would that be?
[tex]\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}[/tex].
Molarity of this strontium hydroxide solutionThere are [tex]8.71467\times 10^{-2}\; \rm mol[/tex] of [tex]\rm Sr(OH)_2[/tex] formula units in this [tex]47\; \rm mL[/tex] solution. Convert the unit of volume to liter:
[tex]V = 47\; \rm mL = 0.047\; \rm L[/tex].
The molarity of a solution measures its molar concentration. For this solution:
[tex]\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
(Rounded to four significant figures.)
In movies about space, there is frequently a space battle scene where ships
explode in big fireballs. Why are these scenes unscientific?
O
A. The explosion would not be a fireball, but would point towards the closest gravity
source, a planet or star. They are only fireballs on Earth because we are already on a
gravity source.
B. None of these
C. There is no oxygen in space, so there can be no combustion.
OD. Space is very cold, there would not be enough heat energy for an explosion to occur.
-
The correct answer is C. There is no oxygen in space, so there can be no combustion.
Explanation:
Fire and flames are the result of a chemical process known as combustion. Moreover, for combustion to occur there are two essential elements. The first one is a fuel or a substance that releases energy and ignites, and the second one is an oxidant, which accepts electrons. This mix and reaction causes high temperatures and release of heat in the form of fire and flames.
This implies, that for fireballs or any other form of fire to exist there must be oxygen or any substance that replaces it. This does not occur in space because the levels of oxygen are extremely low, this means, at least oxygen is added fireballs are not possible in this context as there is no oxygen, and therefore no combustion (Option C).
Answer:
C.) There is no oxygen in space, so there can be no combustion.
Explanation:
I got it correct on founders edtell
Which Group is in the second column of the periodic table?
A. Noble gases
B. Halogens
C. Alkali metals
D. Alkaline earth metals
Answer:
Hey there!
That would be the alkaline earth metals.
Hope this helps :)
Answer: alkaline earth metals
Explanation:
Which of the following metals has a low melting point?
2 A. Rubidium
B. Potassium
C. Calcium
D. Sodium
Answer:
Rubidium
Explanation:
A diode has IS = 10−17 A and n = 1.05. (a) What is the diode voltage if the diode current is 70 μA? (b) What is the diode current for VD = 0.1 mV?
Answer:
(a) The diode voltage, [tex]V_D =[/tex] 0.776 V
(b) The diode current, [tex]I_D =[/tex] 3.81 x 10⁻²⁰ A
Explanation:
Given;
saturation current in diode, [tex]I_s[/tex] = 10⁻¹⁷ A
nonideality factor, n = 1.05
(a) the diode voltage
Given diode current, [tex]I_D[/tex] = 70 μA = 7 x 10⁻⁶ A
Diode voltage is calculated as;
[tex]V_D = nV_Tln(1+ \frac{I_D}{I_S} )[/tex]
Where;
[tex]V_T[/tex] is thermal voltage at 25°C = 0.025
[tex]V_D = 1.05 * 0.025 ln(1+ \frac{70*10^{-6}}{1*10^{-17}})\\\\V_D = 0.02625ln(1+ 7*10^{12})\\\\V_D = 0.776 \ V[/tex]
b) the diode current for VD = 0.1 mV
[tex]V_D = nV_Tln(1 +\frac{I_D}{I_S} )\\\\ln(1 +\frac{I_D}{I_S} ) = \frac{V_D}{nV_T} \\\\ln(1 +\frac{I_D}{I_S} ) = \frac{0.1*10^{-3}}{1.05*0.025} \\\\ln(1 +\frac{I_D}{I_S} ) = 0.00381\\\\1 +\frac{I_D}{I_S} = e^{0.00381}\\\\1+ \frac{I_D}{I_S}= 1.00381\\\\ \frac{I_D}{I_S}=1.00381 - 1\\\\ \frac{I_D}{I_S}= 0.00381\\\\I_D = 0.00381(I_S)\\\\I_D = 0.00381(10^{-17})\\\\I_D = 3.81*10^{-20} \ A[/tex]
when air molecules collide with things around us, it produces _______ (2 words), which is measured with a _______.
✔ When air molecules collide with things around us, it produces pressing force , which is measured with a Pressure gauge.
What is an example of a molecular compound
Answer:
Molecular compounds are inorganic compounds that take the form of discrete (covalent) molecules. Examples include such familiar substances as water (H2O) and carbon dioxide (CO2).
A reaction mixture at 175 K initially contains 522 torr of NO and 421 torr of O2. At equilibrium, the total pressure in the reaction mixture is 748 torr. Calculate Kp at this temperature. Express your answer to three significant figures.
Answer:
[tex]Kp=0.0386[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]
For which the equilibrium expression is:
[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]
Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:
[tex]p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x[/tex]
And the total pressure:
[tex]p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x[/tex]
Yet it is 748 torr, for which the extent is:
[tex]x=943-p_{eq}=943-748\\\\x=195torr[/tex]
Therefore, Kp turns out:
[tex]Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386[/tex]
Best regards.
1. (2 pts) How does the wavelength of a radio wave relate to its frequency? a.) The wavelength gets longer as the frequency increases b.) The wavelength gets shorter as the frequency increases c.) There is no relationship between wavelength and frequency d.) The wavelength depends on the bandwidth of the signal 2. (2 pts) List the following EMR in the order of increasing wavelength starting with the lowest: Infrared radiation Ultraviolet radiation X-rays Visible light 3. (3 pts) Green light has a wavelength of 5.0 x 102 nm. What is the energy, in joules, of ONE photon of green light? What is the energy, in joules of 1.0 mol of photons of green light?
Answer:
1. b.) The wavelength gets shorter as the frequency increases
2. X-rays < Ultraviolet radiation < Visible light < Infrared radiation
3. 2 × 10⁵ J
Explanation:
1. Wavelength vs frequency
fλ= c
f = c/λ
Thus, frequency and wavelength are inversely proportional.
The wavelength increases (gets longer) as the frequency decreases.
2. Order of increasing wavelength
X-rays < Ultraviolet radiation < Visible light < Infrared radiation
3. Energy of green light
(a) Energy of 1 photon
λ = 5 × 10² nm = 5 × 10² × 10⁻⁹ m = 5 × 10⁻⁷ m
fλ = c
f = c/λ = (2.998 × 10⁸ m·s⁻¹)/(5 × 10⁻⁷ m) = 6 × 10¹⁴ s⁻¹
E = hf = 6.626 × 10⁻³⁴ J·s × 6 × 10¹⁴ s⁻¹ = 4 × 10⁻¹⁹ J
(b) Energy of 1.0 mol of photons
[tex]\text{Energy} = \text{1.0 mol photons} \times \dfrac{6.022 \times 10^{23}\text{ photons }}{\text{1 mol photons }} \times \dfrac{4 \times 10^{-19}\text{ J}}{\text{1 photon }} = \mathbf{2 \times 10^{5}} \textbf{ J}\\\\\text{The energy of 1.0 mol of photons of green light is $\large \boxed{\mathbf{2 \times 10^{5}}\textbf{ J}}$}[/tex]
Note: The answer can have only one significant figure because that is all you gave for the wavelength of the light.
A 45.0 mL sample of 0.020 M acetic acid (HC2H3O2) is titrated with 0.020 M NaOH.? Determine the pH of the solution after adding 35.0 mL of any NaOH. (Ka of acetic acid is 1.8 x 10-5) HC2H3O2 (aq) + NaOH (aq) D NaC2H3O2(aq) + H2O (l) (Hint: Calculate new concentration and ICE table)
Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
.02M
CH₃COOH = CH₃COO⁻ + H⁺
C xC xC
Ka = xC . xC / C = x² C
1.8 x 10⁻⁵ = x² . .02
x² = 9 x 10⁻⁴
x = 3 x 10⁻²
= .03
concentration of H⁺ = xC = .03 . .02
= 6 x 10⁻⁴ M , volume = 45 x 10⁻³ L
moles of H⁺ = 6 X 10⁻⁴ x 45 x 10⁻³
= 270 x 10⁻⁷ moles
= 2.7 x 10⁻⁵ moles
concentration of NaOH = .0200 M , volume = 35 x 10⁻³ L
moles of Na OH = 2 X 10⁻² x 35 x 10⁻³
= 70 x 10⁻⁵ moles
=
NaOH is a strong base so it will dissociate fully .
there will be neutralisation reaction between the two .
Net NaOH remaining = (70 - 2.7 ) x 10⁻⁵ moles
= 67.3 x 10⁻⁵ moles of NaOH
Total volume = 45 + 35 = 80 x 10⁻³
concentration of NaOH after neutralisation.= 67.3 x 10⁻⁵ / 80 x 10⁻³ moles / L
= 8.4125 x 10⁻³ moles / L
OH⁻ = 8.4125 x 10⁻³
H⁺ = 10⁻¹⁴ / 8.4125 x 10⁻³
= 1.1887 x 10⁻¹²
pH = - log ( 1.1887 x 10⁻¹² )
= 12 - log 1.1887
= 12 - .075
= 11.925 .
Use your periodic table and calculator as needed for the following question.
How much stock solution is needed to make 250 mL of a 6.0M solution. The molarity of the stock solution is 18M.
Selections may be rounded so choose the best answer.
56 mL
83 mL
2.3 mL
4.7 ml
Medical implants and high-quality jewelry items for body piercings are frequently made of a material known as G23Ti or surgical-grade titanium. The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium. What is the empirical formula for surgical-grade titanium
Answer:
The Empirical Formular is given as; Ti₆Al₄V
Explanation:
The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium.
Elements Titanium Aluminium Vanadium
Percentage 64.39 24.19 11.42
Divide all through by their molar mass
64.39 / 47.87 24.19 / 27 11.42 / 50.94
= 1.345 = 0.896 = 0.224
Divide all though by the smallest number (0.224)
1.345 / 0.224 0.896 / 0.224 0.224 / 0.224
= 6 = 4 = 1
The Empirical Formular is given as; Ti₆Al₄V
Using the stepwise procedure for obtaining the empirical formula of a compound, the empirical formula is [tex] T_{6}Al_{4}V[/tex]
Titanium :
Percentage composition = 64.39%Molar mass = 47.87Divide by Molar mass : = 64.39/47.87 = 1.345
Aluminum :
Percentage composition = 24.19%Molar mass = 27Divide by Molar mass : = 24.19/27 = 0.896
Vanadium :
Percentage composition = 11.42%Molar mass = 50.94%Divide by Molar mass : = 11.42/50.94 = 0.224
Divide by the smallest :
Titanium = 1.345 / 0.224 = 6.00
Aluminum = 0.896 / 0.224 = 4
Vanadium = 0.224 / 0.224 = 1
Hence, the empirical formula is [tex] T_{6}Al_{4}V[/tex]
Learn more : https://brainly.com/question/17091379
For each of the following, classify the substance as a strong acid, strong base, weak acid, or weak base (or perhaps not acidic or basic). Then determine the pH of the solution and calculate the concentrations of all aqueous species present in the solution.a. 2.0 × 10 ^–2 M HBrb. 1.0 × 10^–4 M NaOHc. 0.0015 M Ba(OH)2 d. 0.25 M HCN e. 2.0 × 10 ^–10 M KOH f. 0.050 M NH3 g. 0.100 M NH4Cl h. 0.200 M CaF2 i. 0.0500 M Ba(NO3)2 j. 0.100 M Al(NO3)3
Answer:
a. Strong acid, pH = 1.69
b. Strong base, pH = 10
c. Strong base, pH = 11
d. Weak acid, pH = 4.90
e. Strong base, pH ≅ 7 (pH should be higher than 7, but the base is so diluted)
f. Weak base, pH = 10.96
g. Acidic salt, pH = 5.12
h. Basic salt, pH = 8.38
i. Neutral salt, pH = 7
j. Acidic salt, pH < 7
Explanation:
a. HBr → H⁺ + Br⁻
Hydrobromic acid is a strong acid.
pH = - log [H⁺]
- log 0.02 = 1.69
b. NaOH → Na⁺ + OH⁻
Sodium hydroxide is a strong base.
pH = 14 - pOH
pOH = - log [OH⁻]
pH = 14 - (-log 0.0001) = 10
c. Ba(OH)₂ → Ba²⁺ + 2OH⁻
Barium hydroxide is a strong base
[OH⁻] = 2 . 0.0015 = 0.003M
pH = 14 - (-log 0.003) = 11
d. HCN + H₂O ⇄ H₃O⁺ + CN⁻
This is a weak acid, it reacts in water to make an equilibrium between the given protons and cyanide anion.
To calculate the [H₃O⁺] we must apply, the Ka
Ka = [H₃O⁺] . [CN⁻] / [HCN]
6.2×10⁻¹⁰ = x² / 0.25-x
As Ka is really small, we can not consider the x in the divisor, so we avoid the quadratic formula.
[H₃O⁺] = √(6.2×10⁻¹⁰ . 0.25) = 1.24×10⁻⁵
-log 1.24×10⁻⁵ = 4.90 → pH
e. KOH → K⁺ + OH⁻
2×10⁻¹⁰ M → It is a very diluted concentration, so we must consider the OH⁻ which are given, by water.
In this case, we propose the mass and charges balances equations.
Analytic concentration of base = 2×10⁻¹⁰ M = K⁺
[OH⁻] = K⁺ + H⁺ → Charges balance
The solution's hydroxides are given by water and the strong base.
Remember that Kw = H⁺ . OH⁻, so H⁺ = Kw/OH⁻
[OH⁻] = K⁺ + Kw/OH⁻. Let's solve the quadratic equation.
[OH⁻] = 2×10⁻¹⁰ + 1×10⁻¹⁴ /OH⁻
OH⁻² = 2×10⁻¹⁰. OH⁻ + 1×10⁻¹⁴
2×10⁻¹⁰. OH⁻ + 1×10⁻¹⁴ - OH⁻²
We finally arrived at the answer [OH⁻] = 1.001ₓ10⁻⁷
pH = 14 - (- log1.001ₓ10⁻⁷) = 7
The strong base is soo diluted, that water makes the pH be a neutral value.
Be careful, if you determine the [OH⁻] as - log 2×10⁻¹⁰, because you will obtain as pOH 9.69, so the pH would be 4.31. It is not possible, KOH is a strong base and 4.30 is an acid pH.
f. Ammonia is a weak base.
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Kb = OH⁻ . NH₄⁺ / NH₃
1.74×10⁻⁵ = x² / 0.05 - x
We can avoid the x from the divisor, so:
[OH⁻] = √(1.74×10⁻⁵ . 0.05) = 9.32×10⁻⁴
pH = 14 - (-log 9.32×10⁻⁴ ) = 10.96
g. NH₄Cl, an acid salt. We dissociate the compound:
NH₄Cl → NH₄⁺ + Cl⁻. We analyse the ions:
Cl⁻ does not make hydrolisis to water. In the opposide, the ammonium can react given OH⁻ to medium, that's why the salt is acid, and pH sould be lower than 7
NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺ Ka
Ka = NH₃ . H₃O⁺ / NH₄⁺
5.70×10⁻¹⁰ = x² / 0.1 -x
[H₃O⁺] = √ (5.70×10⁻¹⁰ . 0.1) = 7.55×10⁻⁶
pH = - log 7.55×10⁻⁶ = 5.12
As Ka is so small, we avoid the x from the divisor.
h. CaF₂ → Ca²⁺ + 2F⁻
This is a basic salt.
The Ca²⁺ does not react to water. F⁻ can make hydrolisis because, the anion is the strong conjugate base, of a weak acid.
F⁻ + H₂O ⇄ HF + OH⁻ Kb
Kb = x² / 2 . 0.2 - x
Remember that, in the original salt we have an stoichiometry of 1:2, so 1 mol of calcium flouride may have 2 moles of flourides.
As Kb is small, we avoid the x, so:
[OH⁻] = √(1.47×10⁻¹¹ . 2 . 0.2) = 2.42×10⁻⁵
14 - (-log 2.42×10⁻⁵) = pH → 8.38
i . Neutral salt
BaNO₃₂ → Ba²⁺ + 2NO₃⁻
Ba²⁺ comes from a strong base, so it is the conjugate weak acid and it does not react to water. The same situation to the nitrate anion. (The conjugate weak base, from a strong acid, HNO₃)
pH = 7
j. Al(NO₃)₃, this is an acid salt.
Al(NO₃)₃ → Al³⁺ + 3NO₃⁻
The nitrate anion is the conjugate weak base, from a strong acid, HNO₃ so it does not make hydrolisis. The Al³⁺ comes from the Al(OH)₃ which is an amphoterous compound (it can react as an acid or a base) but the cation has an acidic power.
Al·(H₂O)₆³⁺ + H₂O ⇄ Al·(H₂O)₅(OH)²⁺ + H₃O⁺
Which phase change is an example of an exothermic process?
A.
solid to liquid
B.
solid to gas
C.
liquid to solid
D.
liquid to gas
E.
solid to plasma
Reset
Answer:
C
Explanation:
Turning liquid to a solid is like freezing water to ice and requires the water to LOSE (release) heat causing an exothermic reaction.
Convert the following measurement
Answer:
6.9 Kg/mol•dL
Explanation:
To convert 6.9×10⁴ g/mol•L to kg/mol•dL,
First, we shall convert to kg/mol•L.
This can be achieved by doing the following:
Recall: 1 g = 1×10¯³ Kg
1 g/mol•L = 1×10¯³ Kg/mol•L.
Therefore,
6.9×10⁴ g/mol•L = 6.9×10⁴× 1×10¯³
6.9×10⁴ g/mol•L = 69 Kg/mol•L
Finally, we shall convert 69 Kg/mol•L to Kg/mol•dL.
This is illustrated below:
Recall: 1 L = 10 dL
1 Kg/mol•L = 1×10¯¹ Kg/mol•dL
Therefore,
69 Kg/mol•L = 69 × 1×10¯¹
69 Kg/mol•L = 6.9 Kg/mol•dL
Therefore, 6.9×10⁴ g/mol•L is equivalent to 6.9 Kg/mol•dL.
Calculate the amount of HCl in grams required to react with 3.75 g of CaCO3 according to the following reaction: CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Answer:
The correct answer is 2.75 grams of HCl.
Explanation:
The given balanced equation is:
CaCO₃ (s) + 2HCl (aq) ⇒ CaCl₂ (aq) + H₂O (l) + CO₂ (g)
Based on the given information, one mole of calcium carbonate is reacting with two moles of HCl. The molecular mass of HCl is 36.5 grams, thus, the mass of 2 moles of HCl will be, 36.5 × 2 = 73 grams
The molecular mass of CaCO₃ is 100 gram per mole, that is, the mass of 1 mole of CaCO₃ is 100 grams, therefore, the mass of HCl required for reacting with 3.75 grams of CaCO₃ will be,
= 3.75 × 2 × 36.5 / 100 = 2.74 grams of HCl.
Using the periodic table provided, identify the atomic mass of sodium (Na) . Your answer should have 5 significant figures. Provide your answer below: __ amu
Answer:
Your answer will either be 22.9897 or 22.990 !!
Explanation:
Q1) How much heat is released when 6.38 grams of Ag(s) (m.m = 107.9 g/mol) reacts by the equation shown below at
standard state conditions?
4A9 (s) + 2H,Sq) + O2(g)
2Ag $(s) + 2H200)
Substance
AHof (kJ/mol)
-20.6
H259)
Ag2S (5)
H200
-32.6
-285.8
a)
8.80 KI
b) 69.9 kJ
C) 22.1 kJ
d) 90.8 kJ
e) 40.5 kJ
Answer:
The correct answer is -8.80 kJ.
Explanation:
The ΔH° can be determined by using the formula,
ΔH°rxn = ΔH°f (products) - ΔH°f(reactants)
Based on the given information, the ΔH°f of H2S(g) is -20.6, for Ag2S (s) is -32.6 and for H2O (l) is -285.8 kJ/mole.
Now putting the values we get,
= [2 molΔH°f (Ag2S) + 2 molΔH°f (H2O)] - [4 molΔH°f(Ag) + 2 molΔH°f(H2S) + 1 molΔH°f(O2)]
=[2 mol (-32.6 kJ/mol) + 2 mol(-285.8 kJ/mol)] - [4 mol(0.00 kJ/mol) + 2 mol (-20.6 kJ/mol) + 1 mol (0.00 kJ/mol)
= [(-65.2 kJ) + (-571.6 kJ)] - [(-41.2 kJ)]
= -595.6 kJ
Thus, the enthalpy change of -595.6 kJ takes place when 4 mol of Ag reacts by the equation mentioned.
The mass of Ag given is 6.38 grams, the molecular mass of Ag is 107.9 g/mol. The formula for calculating moles is,
Moles = mass/molar mass
= 6.38 g / 107.9 g/mol
= 0.0591 mol
Now the change in enthalpy when 0.0591 mol of Ag reacts by the given reaction is (-595.6 kJ/4 mol) × 0.0591 mol = -8.80 kJ
The negative sign indicates that the heat is released in the process. Therefore, the -8.80 kJ of heat is released by 6.38 grams of Ag in the given case.
Ag+(aq)+2NH3(aq)⇌Ag(NH3)2+(aq) : A g + ( a q ) + 2 N H 3 ( a q ) ⇌ A g ( N H 3 ) 2 + ( a q ) : blank is the Lewis acid and blank is the Lewis base. is the Lewis acid and A g + ( a q ) + 2 N H 3 ( a q ) ⇌ A g ( N H 3 ) 2 + ( a q ) : blank is the Lewis acid and blank is the Lewis base. is the Lewis base.
Answer:
Silver ion - Lewis acid, Ammonia - Lewis base
Explanation:
The reaction is given as;
Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)]2+(aq)
A lewis acid is an electron pair acceptor. While a lewis base is any substance that that can donate a pair of nonbonding electrons.
This reaction however is a complexation reaction, where ammonia is reacting with the silver ion.
Silver ion accepts electrons in this reaction, hence it is the lewis acid. The ammonia on the other hand donates the electrons used in bonding so it is the lewis base.