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3 years ago

How much potassium chloride is needed to make 0.500 m solution with 1.50 L of water?

Chemistry

1 answer:

Andrew [12]3 years ago

4 0

Answer:

55.9 g KCl.

Explanation:

Hello there!

In this case, according to the definition of molality for the 0.500-molar solution, we need to divide the moles of solute (potassium chloride) over the kilograms of solvent as shown below:

$m=\frac{mol}{kilograms}$

Thus, solving for the moles of solute, we obtain:

$mol=m*kilograms$

Since the density of water is 1 kg/L, we obtain the following moles:

$mol=0.500mol/kg*1.50kg\\\\mol=0.75mol$

Next, since the molar mass of KCl is 74.5513 g/mol, the mass would be:

$0.75mol*\frac{74.5513g}{1mol}\\\\55.9g \ KCl$

Regards!

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What is the average kinetic energy and rms speed of N₂ molecules at STP? Compare these values with those of H₂ molecules at STP.

Otrada [13]

The average kinetic energy and rms speed of N₂ molecules at STP is $5.65686 \times 10^{-21}$ and $493 \mathrm{~m} / \mathrm{s}$

Given,

$\begin{aligned}&\mathrm{P}=1.013 \times 10^{5} \mathrm{~Pa} \\&\mathrm{~T}=273.15 \mathrm{~K} \\&\rho=1.25 \mathrm{~kg} / \mathrm{m}^{3}\end{aligned}$

The average kinetic energy of a molecule is given by, $K . E .=\frac{3}{2} k T$ where k is the Boltzmann constant and Tis the absolute temperature of the gas.

$K . E .=\frac{3}{2} \times $1.380649 \times 10^{-23}$ \times 273.15 K$

$K.E=$5.65686 \times10^{-21}$$

The rms speed of $\mathrm{N}_{2}$ molecules is given by

$\begin{aligned}&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{3 \mathrm{P}}{\rho}} \\&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{P}}{\rho}} \\&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \times 1.013 \times 10^{5}}{1.25}}=493 \mathrm{~m} / \mathrm{s}\end{aligned}$

The average kinetic energy of a gas's particles is inversely related to its temperature. As the gas warms, the particles must travel more quickly since their mass is constant.

The average kinetic energy (K) is equal to one half of the mass (m) of each gas molecule times the RMS speed (vrms) squared.

Learn more about average kinetic energy brainly.com/question/1599923

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2 years ago

6.00g of gold was heated from 20.0c to 22.0c. How much heat was applied to the gold

Alinara [238K]

6 x .129 x 2= 1.55J
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2 years ago

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Answer:

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2 years ago

A large balloon contains 5400 m3 of He gas that is kept at a temperature of 280 K and an absolute pressure of 1.10 x 105 Pa. Fin

patriot [66]

Answer:

1.02 × 10⁶ g

Explanation:

Step 1: Given data

Volume of the balloon (V): 5400 m³

Temperature (T): 280 K

Absolute pressure (P): 1.10 × 10⁵ Pa

Molar mass of He (M): 4.002 g/mol

Step 2: Convert "V" to L

We will use the conversion factor 1 m³ = 1000 L.

5400 m³ × 1000 L/1 m³ = 5.400 × 10⁶ L

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 101325 Pa.

1.10 × 10⁵ Pa × 1 atm / 101325 Pa = 1.09 atm

Step 4: Calculate the moles of He (n)

We will use the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.09 atm × 5.400 × 10⁶ L / 0.08206 atm.L/mol.K × 280 K

n = 2.56 × 10⁵ mol

Step 5: Calculate the mass of He (m)

We will use the following expression.

m = n × M

m = 2.56 × 10⁵ mol × 4.002 g/mol

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8 0

3 years ago

10 points. Please help.

ratelena [41]

Answer:

-191.7°C

Explanation:

P . V = n . R . T

That's the Ideal Gases Law. It can be useful to solve the question.

We replace data:

2.5 atm . 8 L = 3 mol . 0.082 L.atm/mol.K . T°

(2.5 atm . 8 L) / (3 mol . 0.082 L.atm/mol.K) = T°

T° = 81.3 K

We convert T° from K to C°

81.3K - 273 = -191.7°C

6 0

3 years ago

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