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Eddi Din [679]
3 years ago
11

Calculate the percentage by mass of chlorine in Cobalt (II) chloride (cocl2)​

Chemistry
2 answers:
Likurg_2 [28]3 years ago
8 0

Explanation:

Divide the mass of chlorine by the molar mass of cobalt chloride, then multiply by 100.

Molar Mass of Cobalt Chloride.

Mass of Chlorine in Cobalt Chloride.

Percent Composition of Chlorine.

erastovalidia [21]3 years ago
3 0
Divide the mass of chlorine by the molar mass of cobalt chloride, then multiply by 100.
Molar Mass of Cobalt Chloride.
Mass of Chlorine in Cobalt Chloride.
Percent Composition of Chlorine.
Mar 16, 2016
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.One ballon is filled with Hydrogen (H2) gas, the other balloon is filled with methane (CH4). Both balloons have the same temper
Neko [114]

Answer:

C. The balloon with CH4 has the same moles of gas molecules as the balloon with H2

Explanation:

Based on combined gas law, gases under the same pressure, temperature and volume have the same number of moles. With this information we can say the rigth statement is:

<h3>C. The balloon with CH4 has the same moles of gas molecules as the balloon with H2</h3>
7 0
3 years ago
If nitrogen-13 has a half life of 2.5 years, how much remains from a 100g sample after 7.5 years
Sonbull [250]

Answer:

12.50g

Explanation:

T½ = 2.5years

No = 100g

N = ?

Time (T) = 7.5 years

To solve this question, we'll have to find the disintegration constant λ first

T½ = In2 / λ

T½ = 0.693 / λ

λ = 0.693 / 2.5

λ = 0.2772

In(N/No) = -λt

N = No* e^-λt

N = 100 * e^-(0.2772*7.5)

N = 100*e^-2.079

N = 100 * 0.125

N = 12.50g

The sample remaining after 7.5 years is 12.50g

5 0
3 years ago
This unbalanced equation represents a chemical reaction:
inn [45]

Answer:

2

Explanation:

Pb(NO3)2 + 2NaI 2NaNO3 + PbI2

5 0
2 years ago
Read 2 more answers
Draw one product that you would expect from the reaction of 1 mol of 1,3-butadiene and 1 mol of h2o, cat. h2so4?
Daniel [21]
1,3-butadiene is the simplest conjugated diene and undergoes 1,4 addition reaction in acidic environment.
Chemical reaction: CH₂=CH-CH=CH₂ + H₂O → CH₃-CH=CH-CH₂-OH.
CH₂=CH-CH=CH₂ - 1,3-butadiene.
CH₃-CH=CH-CH₂-OH - 2-buten-1-ol.
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3 0
3 years ago
If 15.0 mL of phosphoric acid completely neutralizes 38.5 mL of 0.150 mol/L calcium hydroxide, what is the concentration of the
Sedbober [7]

Answer:

Let me give it a try.

H3PO4 + Ca(OH)2 = Ca3(PO4)2 + H2O

Balancing this reaction

2H3PO4 + 3Ca(OH)2 == Ca3(PO4)2 + 6H2O.

Moles= Molarity x Volume

Volume = 38.5ml = 0.0385L

Moles of Ca hydroxide = 0.150m/L x 0.0385L

(Notice the units canceling out...leaving moles).

=0.005775moles of Ca(OH)2.

From balanced reaction...

3moles of Ca(OH)2 completely reacts with 2moles of H3PO4

0.005775moles of Ca(OH)2 would completely react with....

= 0.005775 x 2/(3)

=0.00385moles of H3PO4.

Now we're looking for its Concentration in Mol/L

Molarity=Moles of solute/Volume of solution(in L)

Volume of solution assuming no other additions to the reaction = 15ml + 38.5ml =53.5ml =0.0535L

Molarity = 0.00385/0.0535

=0.072Mol/L.

If this is wrong

then Simply Try The formula for Mixing of solutions

C1V1 = C2V2

0.15 x 38.5 = C2 x (15+38.5)

C2 = 0.11M/L.

7 0
2 years ago
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