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2 years ago

Please help me with this.And show all work as well ASAP!!

Chemistry

1 answer:

qaws [65]2 years ago

6 0

Answer: The partial pressure of oxygen is 187 torr.

Explanation:

According to Raoult's law, the partial pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the total pressure.

$p_1=x_1p_{total}$

where, x = mole fraction

$p_{total}$ = total pressure

$x_{oxygen}=\frac{\text {moles of oxygen}}{\text {total moles}}$ ,

$x_{oxygen}=\frac{3.0}{12.33}=0.243$ ,

$p_{oxygen}=0.243\times 770torr=187torr$

Thus the partial pressure of oxygen is 187 torr.

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The Ka for acetic acid (HC2H3O2) is 1.80 x 10-5 . Determine the pH of a 0.0500mol/L acetic acid solution.

TiliK225 [7]

Answer:

pH = 3.02

Explanation:

Acetic Acid is a weak acid (HOAc) that ionizes only ~1.5% as follows:

HOAc ⇄ H⁺ + OAc⁻.

In pure water the hydronium ion concentration [H⁺] equals the acetate ion concentration [OAc⁻] and can be determined* using the formula [H⁺] = [OAc⁻] = SqrRt(Ka·[acid]) = SqrRt(1.8x10⁻⁵ x 0.0500)M = 9.5x10⁻⁴M.

By definition, pH = -log[H⁺] = -log(9.5x10⁻⁴) = 3.02

______________________________________________________

*This formula can be used to determine the [H⁺] & [Anion⁻] concentrations for any weak acid in pure water given its Ka-value and the molar concentration of acid in solution.

8 0

3 years ago

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )

Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

$rate = k*[O_{3}][NO]$ (2)

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

$rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s$

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

$\frac{\Delta[NO_{2}]}{\Delta t} = rate$

$\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}$

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!

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3 years ago

Why is peer review an important part of the scientific process?

BigorU [14]

I would say B.

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forsale [732]

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3 years ago

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3 years ago

Please help me with this.And show all work as well ASAP!!​

Please help me with this.And show all work as well ASAP!!