All categories

4 years ago

Q #13 How many moles of MgCl2 are there in 350. g of compound?

Chemistry

1 answer:

Vaselesa [24]4 years ago

8 0

Answer:

3.67 moles

Step-by-step explanation:

We need to find out the number of $MgCl_2$ moles present in 350 grams of a compound.

Molar mass of $Mg$ = 24.305

Molar mass of $Cl_2$ = 35.453

So, one mole of $MgCl_2$ = 24.305 + (35.453 * 2) = 95.211g

1 Mole in 1 molecule of $MgCl_2$ = $\frac{1}{95.211} = 0.0105$

Therefore, number of moles in 350 grams of compound = 0.0105 * 350

= 3.67 moles

You might be interested in

Micheal has a substints that he puts in container 1 the substance has a volume of 5 cubic meters

bazaltina [42]

Answer: Solids and liquids don't suddenly change their volumes.

But gases do...they expand to fill whatever container you

put them in.

Michael's mystery substance is in the gaseous phase.

Explanation:

8 0

3 years ago

Lighting a match is an example of what kind of energy conversion?

Alika [10]

Answer:

Mechanical Energy to Thermal Energy

When you strike a match, it moves through the air until it rubs against a surface. The rubbing produces the heat required to light the match. This is a transformation from mechanical energy to thermal (heat) energy.

Explanation:

4 0

3 years ago

Read 2 more answers

CORRECT ANSWER GETS BRAINLIEST! PLEASE HELP AND EXPLAIN!!

Anna71 [15]

Answer:

A car stopped at the top of the hill

Explanation:

It's potenial energy because, the car could go down the hill and create kinetic energy. Hope this helps! Please give me brainly it is correct!

3 0

3 years ago

A) A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M

Wewaii [24]

Answer:

i) 5.00 mL of 1.00 M NaOH: before the equivalence point

ii) 50.0 mL of 1.00 M NaOH: before the equivalence point

iii) 100 mL of 1.00 M NaOH: at the equivalence point

iv) 150 mL of 1.00 M NaOH: after the equivalence point

v) 200 mL of 1.00 M NaOH: after the equivalence point

Explanation:

1. First calculate the number of mol acid in the 100 mL of 1.00 M HCl solution.

Equation:

Molarity = numbrer of moles of solute / volume of the solution in liters.

Thus, you need to convert each volume from mL to liters, which is done dividing by 1,000.

Naming M the molarity, n the number of moles of solute (acid or base), and V the volume in liters:

$M=n/v\implies n=M\times V=1.00M\times 0.100L=0.100mol$

2. Now calculate the number of moles of NaOH for every condition (addition)

i) 5.00 mL of 1.00 M NaOH

$n=0.00500liter\times 1.00M=0.00500molNaOH$

Since the number of moles of NaOH added (0.00500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

ii) 50.0 mL of 1.00 M NaOH

$n=0.0500liter\times 1.00M=0.0500molNaOH$

Since the number of moles of NaOH added (0.0500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

iii) 100 mL of 1.00 M NaOH

$n=0.100liter\times 1.00M=0.100molNaOH$

Since the number of moles of NaOH added (0.100mol) is equal to the number of moles of acid in the solution (0.100mol), this is at the equivalence point.

iv) 150 mL of 1.00 M NaOH

$n=0.150liter\times 1.00M=0.150molNaOH$

Since the number of moles of NaOH added (0.150mol) is greater than the number of moles of acid in the solution (0.100mol), this is after the equivalence point.

v) 200 mL of 1.00 M NaOH

This is more volume of NaOH, then this is also after the equivalence point.

5 0

3 years ago

Calculate the concentration of an aqueous solution of ca(oh)2 that has a ph of 11.30.

Aleks [24]

Answer:
[OH]- = 501.187 M

Explanation:
First, the given Ca(OH)2 is a base and not an acid. Therefore, we cannot do the calculations based on the pH vale. We need to get the pOH value.
pH + pOH = 14
11.3 + pOH = 14
pOH = 14 - 11.3 = 2.7

Now, pOH is calculated as follows:
pOH = -log [OH]
This means that:
2.7 = - log [OH]
[OH]- = 10^(2.7)
[OH]- = 501.187 M

Hope this helps :)

3 0

4 years ago