Balance the following redox equation using the oxidation-number-change method. Show your work, describing each step.

Hey can someone pls help me answer these questions? It should be easy maybe. It’s just science. The question just won’t to know

miskamm [114]

Hehehehwgwgw. Be the hardest thing ever for a long day and I have a windows

6 0

3 years ago

What is the correct name for the covalent compound N3Se5?

AveGali [126]

Answer:

I think it's Trinitrogen Pentaseleniumide

Explanation:

Tri - three

Penta - five

Second element ends with -ide

6 0

3 years ago

Formic acid, from the Latin formica, is the acid present in ants sting. What is the

leonid [27]

For the reaction of deprotonation of formic acid, the concentration of HCOO⁻ at equilibrium is 0.0151 M if the initial concentration of formic acid is 1.35 M.

The reaction of deprotonation of formic acid is the following:

CHOOH + H₂O ⇄ HCOO⁻ + H₃O⁺

At the equilibrium we have:

CHOOH + H₂O ⇄ HCOO⁻ + H₃O⁺ (1)

1.35 - x x x

The acid equilibrium constant for this reaction is:

$K_{a} = \frac{[HCOO^{-}][H_{3}O^{+}]}{[CHOOH]} = 1.7 \cdot 10^{-4}$ (2)

Entering the values of [CHOOH] = 1.35-x, [HCOO⁻] = [H₃O⁺] = x, into equation (2) we have:

$1.7 \cdot 10^{-4} = \frac{[HCOO^{-}][H_{3}O^{+}]}{[CHOOH]} = \frac{x^{2}}{(1.35 - x)}$

$1.7 \cdot 10^{-4}(1.35 - x) - x^{2} = 0$

After solving the above quadratic equation and taking the positive value for x (concentrations cannot be negative), we have:

$x = [HCOO^{-}] = [H_{3}O^{+}] = 0.0151 M$

Therefore, the concentration of HCOO⁻ at equilibrium is 0.0151 M.

Learn more about the equilibrium constant here:

brainly.com/question/7145687?referrer=searchResults

brainly.com/question/9173805?referrer=searchResults

I hope it helps you!

3 0

3 years ago

3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 1

RUDIKE [14]

Answer:

Enthalpy change for the dissolution of the unknown solid = 4.6 Kj/mole

Explanation:

Using Q = m x Cs x ΔT ................................(1)

where Q = Amount of heat absorbed

m = Mass of solution

Cs = Specific heat capacity of solution

ΔT = Change in temperature

Given Density of solution = 1.20 g/ml

And volume of solution = 150 ml

Mass of solution = density x volume

= 1.2 x 150

= 180 g

From equation (1)

Q = 180 x 4.18 x 19.2 = 14446.08 = 14.4 Kj

So, ΔH of the dissolution of the unknown solid = $\frac{Q}{n}$ = $\frac{14.4}{3.15}$ = 4.6 kj/mole

7 0

3 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

3 years ago