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VikaD [51]
3 years ago
15

We have an aqueous solution that contains 35% (by mass) of a hypothetical solute Z. The formula weight of the solute Z is 118 g/

mol. The density of the solution is observed to be 1.4 g/mL. What is the molarity of Z in this solution
Chemistry
1 answer:
Allisa [31]3 years ago
6 0

Answer:

4.15 M

Explanation:

In order to find the molarity of a stock solution of Z, we use the formula;

Co = 10pd/M

where;

p = percentage by mass of Z

d = density of Z

M= molar mass of Z

Substituting values;

Co= 10 * 35 * 1.4/118

C0= 4.15 M

You might be interested in
If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0
3 years ago
Which of the following can be changed in a chemical reaction?
DanielleElmas [232]

Answer:

Type of molecules

Explanation:

got it right

4 0
3 years ago
Please help ASAP I will give Brainliest
MaRussiya [10]

Answer:

17.6 grams of nitrogen gas

4 0
3 years ago
Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of metha
nirvana33 [79]

Answer:

0.92 kg

Explanation:

The volume occupied by the air is:

35.0m\times 35.0m \times 3.2m \times \frac{10^{3}L }{1m^{3} } =3.9 \times 10^{6} L

The moles of air are:

3.9 \times 10^{6} L \times \frac{1.00mol}{22.4L} =1.7 \times 10^{5}mol

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

1.7 \times 10^{5}mol \times \frac{30J}{K.mol} \times 10.0 K = 5.1 \times 10^{7}J

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:

5.1 \times 10^{7}J \times \frac{1kgCH_{4}}{55.5 \times 10^{6} J } =0.92kgCH_{4}

3 0
3 years ago
What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

3 0
3 years ago
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