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3 years ago

Methylamine (ch3nh2) is a weak base. at equilibrium, which expression equals the base dissociation constant (kb) of methylamine?

1 answer:

4 0

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Below are the choices that can be found elsewhere:

a. [CH3NH-] * [OH-] / [CH3NH3+]
b. [CH3NH3+] / ([CH3NH2] * [OH-]) 
c. [CH3NH3+] * [OH-] / [CH3NH2] 
d. [CH3NH3+] * [OH-] / [CH3NH2] * [H2O]

The answer is C.

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The reaction Ba(NO3)2(aq) + Na2SO4(aq) >> 2NaNO3(aq) + BaSO4(s) goes to completion because a

Helen [10]

Answer:

2. precipitate is formed

Explanation:

In this reaction barium nitrate and sodium sulfate solutions react to form a barium sulfate precipitate.Barium and sulfate ions will react to give barium sulfate precipitate where as the sodium and nitrate ions are spectator ions.

The net ionic equation after removing unchanged ions from each side of the full ionic equation will be;

Ba²⁺+ SO₄²⁻ → BaSO₄ (s)

3 0

3 years ago

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RoseWind [281]

Answer:

i wold think it is C

Explanation:

8 0

3 years ago

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Magnesium reacts with hydrochloric acid (HCl) as follows. Mg(s) + 2HCl(aq) -->MgCl(2)+H(2)(g) How many milliliters of hydroge

max2010maxim [7]

We are given the number of moles of solid magnesium supplied for the reaction which is 0.02 moles while hydrochloric acid is supplied in excess thus we can say that the reaction proceeds to completion. Calculation is as follows:

0.020 mol Mg ( 1 mol H2 / 1 mol Mg ) = 0.020 mol H2 gas is produced

To convert the number of moles to volume, we use the conditions at STP of 1 mol of a substance is equal to 22.4 L. Thus,

0.020 mol H2 (22.4 L / 1 mol) (1000 mL / 1 L) = 448 mL

5 0

4 years ago

The equilibrium constant, Kc, for the reaction N2O4(g)⇌2NO2(g) is 5.0×10−3. If the equilibrium mixture contains [NO2] = 0.050 M

Sergeu [11.5K]

Answer:

0.50 M is the molar concentration of $N_2O_4$ .

Explanation:

The equilibrium constant of the reaction = $K_c=5.0\times 10^{-3}$

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Equilibrium concentration of $NO_2$ = 0.050 M

$[NO_2]=0.050 M$

Equilibrium concentration of $N_2O_4$ = ?

$[N_2O_4]= x$

The equilibrium constant of the reaction is given aSL:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(0.050 M)^2}{x}=5.0\times 10^{-3}$

$x = 0.50 M$

0.50 M is the molar concentration of $N_2O_4$ .

4 0

3 years ago

Calculate ΔHo for the following reaction ussing the given bond dissociation energiesCH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)BOND

Mazyrski [523]

Answer:

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

Explanation:

$CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,$ ΔH° = ?

We are given with:

$\Delta H_{O-O}=142 kJ/mol$

$\Delta H_{O=O}=498 kJ/mol$

$\Delta H_{H-O}=459 kJ/mol$

$\Delta H_{C-H}=411 kJ/mol$

$\Delta H_{C-O}=358 kJ/mol$

$\Delta H_{C=O}=799 kJ/mol$

ΔH° =

(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

$\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})$

$\Delta H^o=(1 mol\times 4\times 411 kJ/mol+2 mol\times 1\times 498 kJ/mol)-(1 mol\times 2\times 799 kJ/mol+2 mol\times 2\times 459 kJ/mol)$

$\Delta H^o=-794kJ$

$\Delta H^o>0$ endothermic reaction

$\Delta H^o$ exothermic reaction

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

4 0

4 years ago