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4 years ago

Determine the quantity (g) of pure CaCl2 in 7.5 g of CaCl2•9H2O. Show your work.

Chemistry

1 answer:

Alisiya [41]4 years ago

3 0

Answer: The quantity of pure $CaCl_2$ in 7.5 g of $CaCl_2.9H_2O$ is 3.0 g

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}=\frac{7.5g}{273g/mol}=0.027moles$

As 1 mole of $CaCl_2.9H_2O$ contains = 1 mole of $CaCl_2$

Thus 0.027 moles of $CaCl_2.9H_2O$ contains = $\frac{1}{1}\times 0.027=0.027$ mole of $CaCl_2$

Mass of $CaCl_2=0.027mol\times 111g/mol=3.0g$

Thus the quantity of pure $CaCl_2$ in 7.5 g of $CaCl_2.9H_2O$ is 3.0 g

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3 years ago

How much heat in joules and in calories is required to heat at 28.4g( 1 oz) ice cube from -23C TO 1.0C

Tom [10]

1426.58 J and 340.90 calories heat in joules and in calories is required to heat at 28.4g( 1 oz) ice cube from -23C TO 1.0C.

Explanation:

Data given:

mass = 28.4 gram

initial temperature = -23 degrees

final temperature = 1degress

change in temperature ΔT = Tfinal - Tinitial

ΔT = 1 -(-23)

ΔT = 24 degrees

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Formula used:

q = mc ΔT

putting the values in the equation:

q= 28.4 x 2.093 x 24

= 1426.58 J

ENERGY IN CALORIES:

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3 years ago

Percocet contains oxycodone at varying strengths.Every strength also contains 325 mg of

Marat540 [252]

5.56 Rhats The right answer

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3 years ago

A 25.0 liter rigid container has a mixture of 32.00 grams of oxygen gas and 1 point

Ksivusya [100]

Answer:

$P_{T}$ = 2.94 atm

Explanation:

The total pressure ( $P_{T}$ ) in the container is given by:

$P_{T} = P_{O_{2}} + P_{He}$

The pressure of the oxygen ( $P_{O_{2}}$ ) and the pressure of the helium ( $P_{He}$ ) can be calculated using the ideal gas law:

$PV = nRT$

<u>Where</u>:

V: is the volume = 25.0 L

n: is the number of moles of the gases

R: is the gas constant = 0.082 Latm/(Kmol)

T: is the temperature = 298 K

First, we need to find the number of moles of the oxygen and the helium:

$n_{O_{2}} = \frac{m}{M}$

Where m is the mass of the gas and M is the molar mass

$n_{O_{2}} = \frac{32.00 g}{31.998 g/mol} = 1.00 moles$

And the number of moles of helium is:

$n_{He} = \frac{8.00 g}{4.0026 g/mol} = 2.00 moles$

Now, we can find the pressure of the oxygen and the pressure of the helium:

$P_{O_{2}} = \frac{nRT}{V} = \frac{1.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 0.98 atm$

$P_{He} = \frac{nRT}{V} = \frac{2.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 1.96 atm$

Finally, the total pressure in the container is:

$P_{T} = P_{O_{2}} + P_{He} = 0.98 atm + 1.96 atm = 2.94 atm$

Therefore, the total pressure in the container is 2.94 atm.

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4 years ago

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julsineya [31]

Answer:

According to my search, the answer is this.

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