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3 years ago

In an aqueous solution at 25°C, if [H30+] = 3.3 * 10^4 M, then [OH-] is:

Chemistry

1 answer:

sleet_krkn [62]3 years ago

3 0

Answer:

[OH-] = 3.0 x 10^-19 M

Explanation:

[H3O+][OH-] = Kw

Kw = 1.0 x 10^-14

[H3O+][OH-] = 1.0 x 10^-14

[OH-] = 1.0 x 10^-14 / 3.3 x 10^4 = 3.0 x 10^-19

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A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

$-w=\frac{dm}{dt}$

w is the mass flow

m is the mass of salt

$-v*C=\frac{dm}{dt}$

v is the volume flow

C is the concentration

$C=\frac{m}{V+(6-3)*L/min*t}$

$-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}$

$-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}$

$-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)$

$-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)$

$m=90kg*[2000L/(2000L+3*L/min*t)]$

a) Initially: t=0

$m=90kg*[2000L/(2000L+3*L/min*0)]=90kg$

b) t=210 min (3.5 hr)

$m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg$

c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

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3 years ago

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Fed [463]

Answer:

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Explanation:

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3 years ago

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siniylev [52]

Answer:

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Calculate the volume of a balloon that can hold 113.4 g of nitrogen dioxide, NO2 gas at STP-

Karolina [17]

Answer:

55.18 L

Explanation:

First we convert 113.4 g of NO₂ into moles, using its molar mass:

113.4 g ÷ 46 g/mol = 2.465 mol

Then we<u> use the PV=nRT formula</u>, where:

P = 1atm & T = 273K (This means STP)

n = 2.465 mol

R = 0.082 atm·L·mol⁻¹·K⁻¹

V = ?

Input the data:

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