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4 years ago

How many moles does 13.3 g of Cu contain?

Chemistry

1 answer:

mariarad [96]4 years ago

3 0

<h2>Hello!</h2>

The answer is:

There are 0.209 moles of Cu in 13.g of Cu

Why?

To calculate how many moles does a sample of any element has, we need to use its atomic mass

We are working with Copper (Cu), so we need to find its atomic mass to calculate how many moles does 13.3 g of Cu contains.

So, calculating we have:

$Cu=63.54\frac{g}{mol}$

We have that there is 1 mol per 63.54 grams of Cu.

Now, converting we have:

$13.3g*\frac{1mol}{63.54g}=0.209moles$

We have that there are 0.209 moles of Cu in 13.g of Cu

Have a nice day!

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which has a higher entropy for the reaction: 2nh3(g)→n2(g) 3h2(g) which has a higher entropy for the reaction: reactants product

krok68 [10]

The one that has a higher entropy for the reaction is products.

<h3>What is entropy?</h3>

Entropy is a measureable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty. It is the measurement of the amount of thermal energy per unit of temperature in a system that cannot be used for productive work. It is a measure of a system's molecular disorder or unpredictability since work is produced by organized molecular motion.

It should bm be noted that the entropy of gas is more than entropy of aqueous which is more than the entropy of liquid and the entropy of solid.

On the product side there are more gas than the reactant side. Therefore, product has more entropy.

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3 0

2 years ago

I really need the answers please!

krek1111 [17]

1) D = 13.6 g / mL

2)ethyl alcohol weighs 158g

3)ρ _copper = 8.9 g $cm^{3}$

Explanation:

D = m / V

=306.0 g / 22.5 mL

D= 13.6 g / mL

density = mass / volume

mass = density × volume

=0.789g /ml × 200.0 ml

M=158g

Ethyl alcohol weighs 158g

ρ (density) = Mass / Volume

ρ _copper = 1896 g / 8.4cm × 5.5cm × 4.6cm

= 1896g / 212.5 $cm^{3}$

ρ _copper=8.9 g $cm^{3}$

4 0

3 years ago

A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and

Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = $99.7^oC$

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

$1atm=760mmHg$

or,

$1mmHg=\frac{1}{760}atm$

As, $1mmHg=\frac{1}{760}atm$

So, $752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm$

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

$K=273+^oC$

As, $K=273+^oC$

So, $K=273+99.7=372.7$

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

$(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole$

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

$\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}$

$\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole$

Thus, the mass of one mole of vapor is 32.0 g/mole

8 0

3 years ago

Does the statement describe a scientific law?

ale4655 [162]

Answer:

I think it's C sorry if I am wrong

8 0

3 years ago

Calculate the density of helium if a balloon with a capacity of 5.00 L holds 0.890 g.

liubo4ka [24]

Answer:

Explanation:

Density is m/V. Also, 1 liter = 1000 $cm^3$ . So, we get 0.890/(5*1000) = $1.78*10^{-4}$ g/cm^3. You can convert this to kg/m^3 as well by multiplying it by 10. Depends which one you want.

5 0

3 years ago