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Vikentia [17]
4 years ago
11

How many Mol are in 4.000 grams of MgCl2 ?

Chemistry
1 answer:
grigory [225]4 years ago
4 0

Answer:

The number of mol is: 0, 042 mol in 4 grams of MgCl2

Explanation:

We calculate the weight of 1 mol of MgCl2:

Weight 1mol of MgCl2= weight Mg + (weight Cl)x  2=

24, 3 grams + 2 x 35, 5 grams = 95, 3 grams/mol MgCl2

95, 3 grams------1 mol MgCl2

4 grams      -------x = (4 grams x1 mol MgCl2)/ 95, 3 grams= 0, 04197 mol MgCl2

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Option A - High frequency, short wavelength.

Explanation:

<em>If the sound is high then the frequency is also high but short-wavelength means lots of waves that always have a high pitch-sound and a high frequency. </em>

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What does fermentation mean?
Cerrena [4.2K]

Answer:

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6 0
3 years ago
Need help !!!!! ASAP
Ksivusya [100]
<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of CO_{2}

n=0.120mol

<h2>Why?</h2>

We are asked to calculate the number of moles of the given gas, also, we  are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

PV=nRT

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

R=62.363\frac{mmHg.L}{mol.K}

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

Temperature(K)=Temperature(C\°) + 273K

So, we are given the following information:

Pressure=760mmHg\\Volume=2.965L\\Temperature=25.5C\°=25.5+273K=298.5K

Now, isolating the number of moles, and substituting the given information, we have:

PV=nRT

n=\frac{PV}{RT}

n=\frac{PV}{RT}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole

Hence, we have that there were produced 0.120 moles of CO_{2}

n=0.120mol

Have a nice day!

7 0
3 years ago
Please help me with this homework
ella [17]

Answer:

Density is the correct answer choice.

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3 years ago
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