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2 years ago

How many miles of MgCI2 are there in 345 g of the compound?

Chemistry

1 answer:

Valentin [98]2 years ago

8 0

MgCl2
Mg=24
Cl=35.5×2 = 71
rmm= 71+24=95g
1 mole = molar mass
1 mole = 95g
1g = 1/95 moles
345g= 1/95 × 345
3.6 moles

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What is 10% of 70 so I kmow

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10 x 70 = 100 x Part, or

700 = 100 x Part

Now, divide by 100 and get the answer:

Part = 700 / 100 = <span>7</span>

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3 years ago

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At one time in the Earth’s history, many people thought the world was flat. Which best explains why the model of the Earth is no

GenaCL600 [577]

Answer:

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Explanation:

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3 years ago

One mole of a metallic oxide reacts with one mole of hydrogen to produce two moles of the pure metal

kenny6666 [7]

Answer:

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Explanation:

Hello!

In this case, according to the given amounts, it is possible to write down the chemical reaction as shown below:

$M_2O+H_2\rightarrow 2M+H_2O$

Which means that the metallic oxide has the following formula: M₂O. Next, we can set up the following proportional factors according to the chemical reaction:

$5.00gM_2O*\frac{1molM_2O}{(2*X+16)gM_2O}*\frac{2molM}{1molM_2O}*\frac{XgM}{1molM}=2.32gM$

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So we solve for x as shown below:

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2 years ago

The table below shows some information about four different elements.

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3 years ago

Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.13 m FeCl3 A. Highest boili

Anestetic [448]

Answer:0.30 m KI ---- A. Highest boiling point

0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point

0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point-C

0.13 m FeCl3---- Lowest boiling point-D

Explanation:

Using the boilng point elevation formula

ΔTb=m* kb *i

where m= molality

kb= elevated boiling point constant( here kb values will be same for all soluton)

i= vant hoff factor = number of ions present in a solution

Using the number of ions and molarity present in a solution as a collagative property, since kb is constant, we can determine which of the species has the highest boiling point.

1.) 0.13 m FeCl3= Fe³⁻ + Cl⁻

i=4

ΔTb=m* kb* i= molarity x number of ionsx Kb= 0.13 x 4= 0.52kb

2) 0.19 m Mg(CH3COO)2 = Mg²⁺ + CH₃COO⁻

i= 3

ΔTb=m* kb* i= molarity x number of ions= 0.19 x 3= 0.57kb

3. 0.30 m KI = K⁺ + I⁻

i= 2

ΔTb=m *kb *= imolarity x number of ions xKb= 0.30x 2= 0.60kb

4. 0.53 m Glucose(nonelectrolyte) =

i= 1 for nonelectroytes

ΔTb=m* kb* i = molarity x number of ionsx Kb= 0.53 x 1= 0.53Kb

therefore,

0.30 m KI ---- A. Highest boiling point

0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point

0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point

0.13 m FeCl3---- Lowest boiling point

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3 years ago