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3 years ago

A solution is made

Chemistry

2 answers:

Juliette [100K]3 years ago

5 0

Answer: 14.3

Explanation:

Anna11 [10]3 years ago

3 0

Answer:

99.3%

Explanation:

The percent by mass of the solute can be expressed as:

% mass = $\frac{MassSolute}{MassSolute+MassSolvent}$ * 100%

And for this problem:

Mass of Solute = Mass of sodium lithium chloride = 29 g

Mass of Solvent = Mass of Water

So to calculate the percent by mass first we need to calculate the mass of water, to do so we use its density (1 g/L):

202 mL is equal to (202/1000) 0.202 L.

Density water = mass water / volume

1 g/L = mass water / 0.202 L

Mass water = 0.202 g

Now we have all the data required to calculate the % mass:

% mass = $\frac{29g}{29g+0.202g}$ * 100 % = 99.3%

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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

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3 years ago

What is the BEST definition for friction?

NARA [144]

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3 years ago

You have a 25.2 L sample of gas at 1.25 atm and 25.0 degrees Celsius. How many moles are present in this gas. For your answer, p

Elenna [48]

Answer:

1.29 mol

Explanation:

This is a direct application of the equation for ideal gases.

$PV=nRT$

Where:

P = pressure = 1.25 atm
V = volume = 25.2 liter
R = Universal constant of gases = 0.08206 atm-liter/K-mol
T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
n = number of moles

Solving for n:

$n=\frac{PV}{RT}$

Substituting:

$n=\frac{1.25atm\times 25.2liter}{0.08206atm-liter/K-mol\times298.15K }\\\\n=1.29mol$

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3 years ago

Limitation for osmosis experiment

taurus [48]

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2 years ago

Calculate the amount of energy required to melt 500 g of ice at 0oc. (δhfus=5.96 kj/mole; δhfus is the energy required to melt i

marta [7]

When you want to melt an ice, you only need the latent energy of fusion, δhfus. We use the given value, then multiply this with the given amount to determine the amount of energy. Since the energy is per mole basis, use the molar mass of ice which is 18 g/mol. The solution is as follows:

ΔH = 5.96 kJ/mol * 1 mol/18 g * 500 g
ΔH = 165.56 kJ

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3 years ago