Answer:
a
Explanation:
its made up of carbon and hydrogen
Determine the number of moles of the anhydrous salt present after heating, assuming that the contents of the aluminum cup after heating are pure anhydrous KAl(SO 4 ) 2 .
Answer:
0.2 moles, assuming weight of dried salt
Explanation:
In order to determine the number of moles, we need to be aware of the mass of the substance in question.
Assuming the mass of the dehydrated [tex]KAl(SO_{4} )_{2}.H_{2} O[/tex] is 50g.
No. of moles = mass of substance/ molar mass of the substance.
= [tex]\frac{50g}{39+27+32*2+16*4*2\\)g/mol}[/tex]
= 0.2 moles moles.
Which of these are elimination reactions? Check all that apply.
CH3OH + CH3COOH → CH3CO2CH3 + H20
C3H7OH → C3H6 + H20
H9C2Br + NaOH → C2H4 + NaBr + H20
Answer:
C3H7OH → C3H6 + H20
Explanation:
If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.
Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.
Answer:
B and C. Just finished my lesson on Edge.
Calculate the frequency (Hz) and wavelength (nm)
of the emitted photon when an electron drops from
the n = 4 to the n=2 level in a hydrogen atom
Answer:
wavelength, λ = 486.6 nm
frequency, f = 6.16 * 10¹⁴ Hz
Explanation:
a. Wavelength
Using the wavelength equation; 1/λ = (1/hc) * 2.18 * 10⁻¹⁸ J * (1/nf² - 1/ni²)
Where nf is the final energy level; ni is the initial energy level; h is Planck's constant = 6.63 * 10⁻³⁴ J.s; c is velocity of light = 3 * 10⁸ m/s
1/λ = 1/(6.63 * 10⁻³⁴ J.s * 3 * 10⁸ m/s) * 2.18 * 10⁻¹⁸ J * (1/2² - 1/4²)
1/λ = 2.055 * 10⁶ m
λ = 4.866 * 10⁻⁷ m
wavelength, λ = 486.6 nm
b. Frequency
Using f = c/λ
f = (3 * 10⁸ m/s) / 4.866 * 10⁻⁷ m
frequency, f = 6.16 * 10¹⁴ Hz
Reduction occurs at which electrode?
Answer:
negative charge electrode
Explanation:
In cathode positive ions are picked up to perform reduction.At the same time negative ions are picked up at anode to get oxidized from electrolyte.
Answer:
The electrode that removes ions from the solution :) a p e x
Which of the following solutions would have the highest pH? Assume that they are all 0.10 M in acid at 25°C. The acid is followed by its Ka value.
a. HCHO2, 1.8 x 10-4
b. HF, 3.5 x 10-4
c. HClO2, 1.1 x 10-2
d. HCN, 4.9 x 10-10
e. HNO2, 4.6 x 10-4
Answer:
[tex]HCN~~Ka=4.9x10^-^1^0[/tex]
Explanation:
In this case, we have to remember the relationship between the Ka value and the pH. We can use the general reaction for any acid with his Ka value expression:
[tex]HA~->~H^+~+~A^-[/tex] [tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
In the Ka expression, we have a proportional relationship between Ka and the concentration of [tex]H^+[/tex]. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher
So, if we have to find the higher pH value we need to search the smaller Ka value in this case [tex]HCN~~Ka=4.9x10^-^1^0[/tex].
I hope helps!
HCN has the highest pH among all the acids listed in the question.
The Ka is called the acid dissociation constant. It shows the extent to which an acid is ionized in water. The pH shows the hydrogen ion concentration of water. The higher the Ka, the higher the hydrogen ion concentration and the lower the pH.
Hence, HCN has the lowest Ka and the lowest hydrogen ion concentration. Therefore, HCN has the highest pH among all the acids listed in the question.
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question attached ! asap please
Answer:
A. copper is highly water soluble. It will turn into 5 different hydrates as it absorbs more and more water.
b. Glycerol is easily soluble in water, due to the ability of the polyol groups to form hydrogen bonds with water molecules
c. octane is considered to be non-polar, it will not be soluble in water, since water is a polar solvent. This will happen because octane (hydrocarbons in general) contains neither ionic groups, nor polar functional groups that can interact with water molecules.
d. Nitric acid decomposes into water, nitrogen dioxide, and oxygen, forming a brownish yellow solution.
e. Barium carbonate is a white powder. It is insoluble in water and soluble in most acids
Explanation:
Answer all to the best of your ability please
Answer:
2–Ethyl–3–methlypentanal.
Explanation:
To name the compound given in the question above, we must observe the following:
1. The functional group of the compound is Alkanal i.e Aldehyde,
—CHO and it is located at carbon 1.
Note: the aldehyde functional group is always at carbon 1 and there will be no need to state it's position in the compound.
2. The longest continuous carbon chain is 5 i.e pentane. But the presence of the functional group will replace the –e at the end of pentane with –al, making the name to the pentanal.
3. The substituents attached are:
a. Ethyl, CH2CH3 at carbon 2.
b. Methyl, CH3 at carbon 3.
4. Combine the above to get the name of the compound.
Therefore, the name of the compound is:
2–Ethyl–3–methlypentanal.
An analytical laboratory balance typically measures mass to the nearest 0.1 mg. You may want to reference (Page) Section 21.6 while completing this problem. Part A What energy change would accompany the loss of 0.1 mg in mass
Answer:
The energy change is [tex]E = 9.0 *10^{9}\ J[/tex]
Explanation:
From the question we are told that
Mass loss is [tex]m_l = 0.1 \ mg = 0.1 *10^{-3} mkg = 0.1 *10^{-6} \ kg[/tex]
Generally the energy change that would accompany this loss is mathematically represented as
[tex]E = m * c^2[/tex]
Where c is the speed of light with values [tex]c = 3.0*10^{8} \ m/s[/tex]
[tex]E = 0.1 *10^{-6} * [3.0 *10^{8}]^2[/tex]
[tex]E = 9.0 *10^{9}\ J[/tex]
Identify the Lewis acid and Lewis base from among the reactants in each of the following equations. Match the words in the left column to the appropriate blanks in the sentences on the right.
1. Fe3+ (aq)+6CN (aq) Fe(CN) (aq)______is the Lewis acid and_____is the Lewis base. is the Lewis
2. CI- (aq) + AlCl3 (aq) AlCl4-____is the Lewis acid and______is the Lewis base.
3. AlBr3 + NH3 H3NAlBr3______is the Lewis acid and______is the Lewis base.
A. AlCl3
B. CN-
C. AlBr3
D. Cl-
E. NH3
F. Fe3+
Answer:
1. Lewis acid: F. Fe₃⁺, Lewis base: B. CN⁻
2. Lewis acid: A. AlCl₃, Lewis base: D. Cl⁻
3. Lewis acid: C. AlBr₃, Lewis base: E. NH₃
Hope this helps.
The Lewis acid is chemical substance which possesses an empty orbital and accepts an electron pair from a Lewis base ( donor ), in order to create a Lewis adduct ( molecule created from the bonding of Lewis base and acid ).
The Lewis acid from reaction 1 is Fe₃⁺ while the Lewis base is CN⁻ also the Lewis acid from reaction 2 is AICI₃ while the Lewis base is CI⁻
Hence we can conclude that the Lewis acids and Lewis bases of the reactions in the question are as listed above.
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Methanol is produced industrially by catalytic hydrogenation of carbon monoxide according to the following equation: CO(g) + 2 H2(g) → CH3OH(l) If the yield of the reaction is 40%, what volume of CO (measured at STP) would be needed to produce 1.0 × 106 kg CH3OH?
Answer:
1.7 × 10⁹ L
Explanation:
Step 1: Write the balanced equation
CO(g) + 2 H₂(g) → CH₃OH(l)
Step 2: Calculate the moles corresponding to 1.0 × 10⁶ kg CH₃OH
The molar mass of CH₃OH is 32.04 g/mol.
[tex]1.0 \times 10^{6} kg \times \frac{10^{3}g }{1kg} \times \frac{1mol}{32.04g} = 3.1 \times 10^{7} mol[/tex]
Step 3: Calculate the theoretical yield of CH₃OH
The real yield of CH₃OH is 3.1 × 10⁷ mol and the percent yield is 40%. The theoretical yield is:
[tex]3.1 \times 10^{7} mol (R) \times \frac{100mol(T)}{40mol(R)} = 7.8 \times 10^{7}mol(T)[/tex]
Step 4: Calculate the moles of CO required to produce 7.8 × 10⁷ mol of CH₃OH
The molar ratio of CO to CH₃OH is 1:1. The moles of CO required are 1/1 × 7.8 × 10⁷ mol = 7.8 × 10⁷ mol
Step 5: Calculate the volume of 7.8 × 10⁷ mol of CO at STP
The volume of 1 mole of CO at STP is 22.4 L.
[tex]7.8 \times 10^{7}mol \times \frac{22.4L}{mol} = 1.7 \times 10^{9}L[/tex]
True or False: Adding 4.18 joules to water will increase the temperature more than adding 1 calorie to water.
Answer:
Because one calorie is equal to 4.18 J, it takes 4.18 J to raise the temperature of one gram of water by 1°C. In joules, water's specific heat is 4.18 J per gram per °C. If you look at the specific heat graph shown below, you will see that 4.18 is an unusually large value.
Explain why o-vanillin does not fully protonate p-toluidine. Reference appropriate pKa values and include a balanced chemical reaction and an appropriate reaction arrow in your answer.
Answer:
Here's what I get
Explanation:
pKₐ of o-vanillin = 7.81; pKₐ of p-toluidine = 4.44
The higher the pKₐ, the weaker the acid.
Thus, o-vanillin is the weaker acid and has a stronger conjugate base.
The conjugate acid of p-toluidine is the stronger and has the weaker conjugate base.
The equation for the equilibrium is
H-OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₂ ⇌ ⁻OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₃⁺
weaker acid weaker base stronger base stronger acid
The reaction between the stronger acid and the stronger base pushes the position of equilibrium to the left.
Thus, o-vanillin does not fully protonate p-toluidine.
O-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base; hence, o-vanillin does not fully protonate p-toluidine.
The pKa is defined as the negative logarithm of Ka. The dissociation constant of an acid Ka shows the extent of dissociation of an acid in solution. The higher the pKa, the lower the Ka and the weaker the acid.
The pKₐ of o-vanillin is 7.81 while the pKₐ of p-toluidine is 4.44. This means that o-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base. Hence, o-vanillin does not fully protonate p-toluidine.
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Assume that a nickel weighs exactly 5.038650 g for the sets of weights listed below obtained by a single weighing on the balance below
Answer:
afshkkyfugutuiryfyi
What states can electrons exist in? A. Electron clouds or energy levels B. Positive and negative C. Up and down spin D. In phase and out of phase
Answer:
A. Electron clouds or energy levels
Explanation:
Electrons can exist in two states:
Stablized in electronic orbitalsFreely movingElectrons can exist in an electron cloud or energy level. Electron in an atoms have ability to change energy levels either by emitting or absorbing a photon that form the energy equal to the energy difference between the two levels.
Hence, the correct answer is A.
Answer:
Up and DOWN spin
Explanation:
How does a balanced chemical equation show the conservation of mass?
A. It shows that the number of each type of atom stays the same.
B. It shows that the mass of the products is greater than the mass of
the reactants when a reaction increases the moles of substances.
C. It shows that the total number of moles of substances stays the
same.
D. It shows that the mass of the reactants is greater than the mass
Answer:
A. It shows that the number of each type of atom stays the same.
Explanation:
Though you may see a change in the way they are arranged, the same number of atoms are present before and after. Balanced chemical equations show equal numbers of atoms of each element on each side of the equation.
The free energy obtained from the oxidation (reaction with oxygen) of glucose (C6H12O6) to form carbon dioxide and water can be used to re-form ATP by driving the above reaction in reverse. Calculate the standard free energy change for the oxidation of glucose.
Answer:
The correct answer is -2878 kJ/mol.
Explanation:
The reaction that takes place at the time of the oxidation of glucose is,
C₆H₁₂O₆ (s) + 6O₂ (g) ⇒ 6CO₂ (g) + 6H₂O (l)
The standard free energy change for the oxidation of glucose can be determined by using the formula,
ΔG°rxn = ∑nΔG°f (products) - ∑nΔG°f (reactants)
The ΔG°f for glucose is -910.56 kJ/mol, for oxygen is 0 kJ/mol, for H2O -237.14 kJ/mol and for CO2 is -394.39 kJ/mol.
Therefore, ΔG°rxn = 6 (-237.14) + 6 (-394.39) - (-910.56)
ΔG°rxn = -2878 kJ/mol
Q1. Calculate the amount of copper produced in 1.0 hour when aqueous CuBr2 solution was electrolyzed by using a current of 4.50 A. Q2. In another electroplating experiment, if electric current was passed for 3 hours and 2.00 g of silver was deposited from a AgNO3 solution, what was the current used in amperes
Answer:
[tex]\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}[/tex]
Explanation:
Q1. Mass of Cu
(a) Write the equation for the half-reaction.
Cu²⁺ + 2e⁻ ⟶ Cu
The number of electrons transferred (z) is 2 mol per mole of Cu.
(b) Calculate the number of coulombs
q = It
[tex]\text{t} = \text{1.0 h} \times \dfrac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} \times \text{ 3600 s} = \textbf{10 800 C}[/tex]
(c) Mass of Cu
We can summarize Faraday's laws of electrolysis as
[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\& = &\dfrac{10 800 \times 63.55}{2 \times 96 485}\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}[/tex]
Note: The answer can have only two significant figures because that is all you gave for the time.
Q2. Current used
(a) Write the equation for the half-reaction.
Ag⁺ + e⁻ ⟶ Ag
The number of electrons transferred (z) is 1 mol per mole of Ag.
(a) Calculate q
[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\2.00& = &\dfrac{q \times 107.87}{1 \times 96 485}\\\\q &=& \dfrac{2.00 \times 96485}{107.87}\\\\& = & \textbf{1789 C}\\\end{array}[/tex]
(b) Calculate the current
t = 3 h = 3 × 3600 s = 10 800 s
[tex]\begin{array}{rcl}q&=& It\\1789 & = & I \times 10800\\I & = & \dfrac{1789}{10800}\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}[/tex]
Note: The answer can have only one significant figure because that is all you gave for the time.
Which of the following metals is paramagnetic?
A. Magnesium
B. Sodium
C. Beryllium
D. Calcium
Answer:
sodium
Explanation:
(Na) atom is paramagnetic and sodium is a na atom.
Nitrogen has different oxidation states in the following compounds: nitrite ion, nitrous oxide, nitrate ion, ammonia, and nitrogen gas. Arrange these species in order of increasing nitrogen oxidation state. Select the correct answer below: A. ammonia, nitrogen gas, nitrite, nitrous oxide, nitrate B. nitrogen gas, ammonia, nitrous oxide, nitrite, nitrate C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate D. ammonia, nitrogen gas, nitrate, nitrite, nitrous oxide
Answer:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
Explanation:
To establish the oxidation number of nitrogen in each compound, we know that the sum of the oxidation numbers of the elements is equal to the charge of the species.
Nitrite ion (NO₂⁻)
1 × N + 2 × O = -1
1 × N + 2 × (-2) = -1
N = +3
Nitrous oxide (NO)
1 × N + 1 × O = 0
1 × N + 1 × (-2) = 0
N = +2
Nitrate ion (NO₃⁻)
1 × N + 3 × O = -1
1 × N + 3 × (-2) = -1
N = +5
Ammonia (NH₃)
1 × N + 3 × H = 0
1 × N + 3 × (+1) = 0
N = -3
Nitrogen gas (N₂)
2 × N = 0
N = 0
The order of increasing nitrogen oxidation state is:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Answer:
V2 = 17371.43ml
Explanation:
We use Boyles laws
since temperature is constant
P1V1=P2V2
760 x 400 = 17.5 x V2
304000 = 17.5 x V2
V2 = 304000/17.5
V2 = 17371.43ml
The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at 760 torrs will be 18 ml.
What is vapor pressure?
The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.
The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -
P1 V1 / T1 = P2 V2 / T1
here, P = pressure
T = temperature
V = volume
substituting the value in the equation,
400 ×760 / 20 = 17.5× V / 20
V = 400× 760 / 20 × 17.5 / 20
V = 18 ml
Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.
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p32p32 is a radioactive isotope with a half-life of 14.3 days. if you currently have 63.163.1 g of p32p32 , how much p32p32 was present 8.008.00 days ago
Answer:
92.93 g
Explanation:
Number of half lives that have elapsed in eight days =8/14.3 = 0.559
Fraction of the radioactive nuclide that remains after 0.559 half lives is given by
N/No=(1/2)^0.559
Where N= mass of radioactive nuclides remaining after a time t
No= mass of radioactive nuclides originally present
N/No=(1/2)^0.559= 0.679
Mass of nuclides present eight days before= 63.1g/0.679
Mass of nuclides present eight days before=92.93 g
1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the actual average molarity you measured in the standardization. What do you notice
Answer:
0.1M solution of NaOH
Explanation:
1 mole of NaOH - 40g
? moles - 1 g = 1/40 = 0.025 moles.
Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume
= 0.025/0.25
= 0.1M.
Interpret the following equation for a chemical reaction using the coefficients given:
Cl2(g) + F2(g) 2ClF(g)
On the particulate level:
________ of Cl2(g) reacts with ______ of F2(g) to form______ of ClF(g).
On the molar level:
______ of Cl2(g) reacts with______ of F2(g) to form______ of ClF(g).
Answer and Explanation:
Given the following chemical equation:
Cl₂(g) + F₂(g) ⇒ 2ClF(g)
The coefficients are: 1 for Cl₂, 1 for F₂ and 2 for ClF. The coefficients indicate the number of units of each ompound that participates in the reaction. It gives the proportion of reactants and products in the reaction. These units can be molecules or moles. In this reaction, we can say:
On the particulate level: 1 molecule of Cl₂(g) reacts with 1 molecule of F₂(g) to form 2 molecules of ClF(g).
On the molar level: 1 mol of Cl₂(g) reacts with 1 mol of F₂(g) to form 2 mol of ClF(g).
oxygen get stable configuration by ____________two electrons
please give the answer as fast as you can
please
Answer:
gaining two electrons
Explanation:
electron configuration
2:6
so add two to 6 to get stable 2:8
Sample gas has a volume of 3.40 L at 10°C what will be its volume at 100°C pressure remaining constant
Answer:
V2 = 4.48L
Explanation:
using charles law
V1/T1=V2/T2
3.4/283=V2/373
0.012=V2/373
V2= 0.012 x 373
V2 = 4.48L
What's the electron configuration of a Ca+2 ion?
A. [Kr]
B. [Ar]
C. [Ne]
D. [He]
Answer:
B
Explanation:
The Ca+2 ion means that 2 electrons have been given away. So when you try and find the answer, you have to count backwards from Calcium. When you do, you get K+ first and then Argon which is either column 8 orc column 18 depending on your periodic table.
The element you hit is Argon.
The answer is B
Answer:
B ar
Explanation:
pen foster answer
Aluminum and oxygen react according to the following equation: 4Al + 3O2 -> 2Al2O3 In a certain experiment, 4.6g Al was reacted with excess oxygen and 6.8g of product was obtained. What was the percent yield of the reaction?
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
4 moles of Al produce 2 moles of Al₂O₃
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.0852 moles of Al₂O₃,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
8.7g of Al₂O₃ can be produced (Theoretical yield)Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
78.2%The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. Assume that the volume of the solutions are additive . What would be the Ka for NH4
Answer:
Following are the answer to this question:
Explanation:
The value of pH solution is =5.17 So, the p^{OH}:
[tex]p^{OH}[/tex]=14-56.17
=8.823
The volume of the [tex]NH_{3}[/tex] = 40.00 ml
convert into the liter= 0.040L
The value of the concentrated [tex]NH_{3}[/tex] =0.10 M
The volume of the [tex]NH_{4}Cl[/tex]= 50.00 ml
convert into the liter= 0.050L
The value of concentrated [tex]NH_{4}Cl[/tex]= 0.10 M
The volume of the [tex]H_{2}So_{4}[/tex]= 30 ml
convert into the liter= 0.030L
The value of concentrated [tex]H_2So_4[/tex]=0.05 M
Calculating total volume=(0.40+0.050+0.030)
=0.120 L
calculating the new concentrated value of [tex]NH_3[/tex] = [tex]\frac{0.10\times 0.040}{0.120}= 0.33 \ M[/tex]
calculating the new concentrated value of [tex]NH_4Cl[/tex]= [tex]\frac{0.050\times 0.10}{0.120}= 0.04166 \ M[/tex]calculating the new concentrated value of [tex]H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M[/tex] when 1 mol [tex]H_2So_4[/tex] produced 2 mols [tex]H^{+}[/tex] so, 0.0125 in [tex]H_2So_4[/tex]produced:
[tex]=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}[/tex]
create the ICE table:
[tex]NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}[/tex]
I (m) 0.033(m) 0.025 0.04166
C -0.025 -0.025 + 0.025
E 8.3\times 10^{-3} 0 0.0667
now calculating pH:
when ph= 8.83:
[tex]P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769[/tex]
Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of copper(II) cation in the solution. You can assume the volume of the solution doesn't change when the copper(II) acetate is dissolved in it. Round your answer to significant digit.
Answer:
Molarity Cu²⁺ = 0.423M Cu²⁺
Explanation:
40.8g of copper (II) acetate into 200mL of a 0.700M sodium chromate
The reaction of copper acetate with sodium chromate occurs as follows:
Cu(CH₃COO)₂(aq) + Na₂CrO₄(aq) → CuCrO₄(s) + 2CH₃COONa
In water, the Copper(II) acetate dissociates in Cu²⁺ cation.
To know final molarity of Cu²⁺ we need to calculate the moles of Cu²⁺ that don't react with chromate ion, thus:
Moles of 40.8g of copper(II) acetate (Molar mass: 181.63g/mol) are:
40.8g × (1mol / 181.63g) = 0.2246 moles of Copper(II) acetate
Moles of sodium chromate are:
0.200L ₓ (0.700mol / L) = 0.140 moles of sodium chromate.
As 1 mole of Copper(II) acetate reacts per mole of sodium chromate, moles of Copper(II) acetate = Moles of Cu²⁺ that remains after the reaction are:
0.2246mol - 0.140moles = 0.0846 moles of Cu²⁺
Molarity is ratio between moles of solute (Moles Cu²⁺) and volume in liters of solution (200mL = 0.200L):
Molarity Cu²⁺ = 0.0846 moles / 0.200L
Molarity Cu²⁺ = 0.423M Cu²⁺Calculate the amount of heat energy, in kJ, needed to increase the temperature of 200g of water by 300 C.
Answer:
Q = 246 kJ
Explanation:
It is given that,
Mass of water, m = 200 g
Let initial temperature, [tex]T_i=5^{\circ}[/tex]
Final temperature of water, [tex]T_f=300^{\circ} C[/tex]
We know that the specific heat capacity of water, [tex]c=4.18\ J/g-^{\circ} C[/tex]
So, the heat energy needed to raise the temperature is given by :
[tex]Q=mc\Delta T\\\\Q=200\times 4.18\times (300-5)\\\\Q=246620\ J[/tex]
or
Q = 246 kJ
So, the heat energy of 246 kJ is needed.