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gogolik [260]
3 years ago
7

Given the following unbalanced equation:

Chemistry
1 answer:
pshichka [43]3 years ago
4 0

Answer:

11.84 moles

Explanation:

We first balance the chemical equation

CoCl_{2} +  F_{2}  ------>   CoF_{2} + Cl_{2}

Both number of elements in the reactant side is equal to elements on the product side thus this equation is already balanced

Using mole ratio to find moles of COF_{2}

1 mole of CoCl2 reacts with 1 mole of F2 to produce 1 mole of CoF2 and ! mole of Cl2

Mole CoCl2 = Mole CoF2 = 11.84 moles

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A solution is prepared by dissolving 23.7 g of cacl2 in 375 g of water. the density of the resulting solution is 1.05 g/ml. the
alekssr [168]
<span>The density of the solution =1.05 g/ml.
</span><span>The total mass of the resulting solution is = 398.7 g (CaCl2 + water)
</span>
Find moles of CaCl2 and water.
Molar mass of CaCl2 = 110 (approx.)

Moles of CaCl2 = 23.7 / 110  = 0.22
so, moles of Cl- ion = 2 x 0.22 = 0.44  (because each molecule of CaCl2 will give two Cl- ions)
Moles of water = 375 / 18 = 20.83

Now, Mole fraction of CaCl2 =  (moles of CaCl2) / (total moles)

total moles = moles of Cl- ions + moles of Ca2+ ions + moles of water

                  = 0.44 + 0.22 + 20.83

=21.49


So, mole fraction = 0.44 / (21.49) = 0.02

Guess what !!! density is not used. No need

4 0
3 years ago
Read 2 more answers
Can someone plz help me? :(
Sveta_85 [38]

Answer:

it's C

Explanation:

because it exhaled the carbon dioxide

8 0
2 years ago
Which of the following elements is not diatomic?
KATRIN_1 [288]

Answer:

Neon

Explanation:

Elements from group 8A stay alone

8 0
3 years ago
A 5 cm3 piece of aluminum has a higher density than a 10cm3 of aluminum. True or false
Lubov Fominskaja [6]

Density stays the same, false

3 0
2 years ago
The normal freezing point of water (H2O) is 0.00 oC and its Kf value is 1.86 oC/m. A nonvolatile, nonelectrolyte that dissolves
adelina 88 [10]

Answer:

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

Explanation:

One colligative property is the freezing point depression due the addition of a solute. The equation is:

ΔT=Kf*m*i

<em>Where ΔT is change in temperature = 0.400°C</em>

<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>

<em>m is molality of the solution (Moles of solute / kg of solvent)</em>

<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>

Replacing:

0.400°C =1.86°C/m*m*1

0.400°C / 1.86°C/m*1 = 0.215m

As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:

0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.

The mass of ethylene glycol must be added is:

0.0602 moles * (62.10g / mol) =

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

<em />

6 0
3 years ago
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