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2 years ago

What is the correct expression for molarity

Chemistry

2 answers:

ki77a [65]2 years ago

8 0

The correct expression for molarity is mmol<span> solute/ml solution. Hope this helps!</span>

Oliga [24]2 years ago

7 0

It is a number of moles of solute divided by the number of liters of solution! Hope this helps! :D

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A solution is prepare using 45.2 g of an unknown (nonelectrolyte) dissolved in 476.4 g of water. The freezing point depression

Ray Of Light [21]

Answer:

94.4g/mol is molar mass of the unknown

Explanation:

Based on the freezing point depression equation:

ΔT = Kf*m*i

<em>Where ΔT is the depression in freezing point (1.87°C)</em>

<em>Kf is freezing point depression constant of water (1.86°Ckg/mol)</em>

<em>And i is Van't Hoff factor (1 for nonelectrolyte solutes)</em>

<em />

Replacing:

1.87°C = 1.86°CKg/mol*m*i

1.005mol/kg solvent = m

Using the mass of the solvent we can find the oles of the nonelectrolyte:

1.005mol/kg solvent * 0.4764kg = 0.479moles

Molar mass is defined as the ratio between mass of a substance in grams and moles, that is:

45.2g / 0.479mol =

<h3>94.4g/mol is molar mass of the unknown</h3>

6 0

2 years ago

If the cylindrical pistons are 25.000 cm in diameter at 20.0 ∘c, what should be the minimum diameter of the cylinders at that te

attashe74 [19]

Refer to the diagram shown below.

The piston supports the same load W at both temperatures.
The ideal gas law is
$pV=nRT$
where
p = pressure
V = volume
n = moles
T = temperature
R = gas constant

State 1:
T₁ = 20 C = 20+273 = 293 K
d₁ = 25 cm piston diameter

State 2:
T₂ = 150 C = 423 K
d₂ = piston diameter

Because V, n, and R remain the same between the two temperatures, therefore
$\frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}}$

If the supported load is W kg, then
$p_{1} = \frac{W \, N}{ \frac{\pi}{4} d_{1}^{2}} = \frac{4W \, N}{\pi (0.25 \, m)^{2}} = 20.3718W \, Pa$
Similarly,
$p_{2} = \frac{4W}{\pi d_{2}^{2}} \, Pa$

$\frac{p_{1}}{p_{2}} = \frac{20.3718 \pi d_{2}^{2}}{4} = 16 d_{2}^{2}$

Because p₁/p₂ = T₁/T₂, therefore
$16d_{2}^{2} = \frac{293}{423} \\\\ d_{2}^{2} = \frac{0.6927}{16} \\\\ d_{2} = 0.2081 \, m$

The minimum piston diameter at 150 C is 20.8 cm.

Answer: 20.8 cm diameter

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3 years ago

How many valence electrons do the noble gases possess?

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Answer:

Have no valence electrons because they are stable.

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Answer:

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Explanation:

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