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4 years ago

Using the equation MgCl2 -> Mg + Cl2, if 5.00 grams of MgCl2 is given, how many grams of Mg is produced?

Chemistry

1 answer:

motikmotik4 years ago

4 0

Answer:

MgCl2-> Mg+Cl

mass 5.00g Mass=1.263g

RFM Mg=24+35.5 ×2= 95 RFM= 24

moles= 5.00÷ 95= 0.0526 Moles=0.0526

Explanation:

mas of Mg= 1.263 grams

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Is the bond C=O polar or non polar

rewona [7]

Answer:

Non Polar

Explanation:

Non Polar molecule because of its linear symmetric shape

6 0

3 years ago

The mole fraction of a non-electrolyte (MM 40.0 g/mol) in a saturated aqueous solution is 0.310. What is the molality of the sol

jeka57 [31]

<u>Answer:</u> The molality of non-electrolyte is 24.69 m

<u>Explanation:</u>

We are given:

Mole fraction of saturated aqueous solution = 0.310

This means that 0.310 moles of non-electrolyte is present.

Moles of water (solvent) = 1 - 0.310 = 0.690 moles

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of water = 0.690 moles

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$0.690mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.690mol\times 18g/mol)=12.42g$

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}$

Where,

$n_{solute}$ = Moles of solute (non-electrolyte) = 0.310 moles

$W_{solvent}$ = Mass of solvent (water) = 12.42 g

Putting values in above equation, we get:

$\text{Molality of non-electrolyte}=\frac{0.310\times 1000}{12.42}\\\\\text{Molality of non-electrolyte}=24.96m$

Hence, the molality of non-electrolyte is 24.69 m

4 0

3 years ago

Convert 100mL to L. Show your work to receive full credit.

Ilya [14]

Answer:

100 ml = 0.1 L

I divided 100 by 1000 because 1000ml = 1L

3 0

3 years ago

When do you stop writing your electron configurations?

Mila [183]

When all the electrons are placed in their orbitals!!!

6 0

3 years ago

Chlorine has two stable isotopes, Cl-35 and Cl-37. If their exact masses are 34.9689 amu and 36.9695 amu, respectively, what is

natima [27]

Answer:

X(Cl-35) = 75.95% => Answer 'A'

Explanation:

34.9689·X(Cl-35) + 36.9695·X(Cl-37) = 35.45; X = fractional abundance

X(Cl-35) + X(Cl-37) = 1 ⇒ X(Cl-37) = 1 - X(Cl-25)

34.9689·X(Cl-35) + 36.9695(1 - X(Cl-35)) = 35.45

34.9689·X(Cl-35) + 36.9695 - 36.9695·X(Cl-35) = 35.45

Rearrange ...

36.9695·X(Cl-35) - 34.9689·X(Cl-35) = 36.9689 - 35.45

2.0006·X(Cl-35) = 1.5195

X(Cl-35) = 1.5195/2.0006 = 0.7595 fractional abundance

⇒ % abundance = 75.95%

3 0

3 years ago

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