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aivan3 [116]
3 years ago
12

15 g NiO is dissolved into enough water to make 800. mL of solution. What is the molar concentration of the solution? Report you

r answer with the correct number of significant figures.
Chemistry
1 answer:
Nuetrik [128]3 years ago
6 0

Answer: 0.25

Explanation: M = moles of solute / volume of solution (L)

First, you need to calculate the number of moles using the given mass and the molar mass.

15 g74.692 g/mol≈0.20 mol

Then, you can solve for molarity using the molarity equation. The answer should be reported to two significant figures.

M=0.20 mol/ 0.800 L = 0.25 M

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Which of the following are classified as an electron group? A.Charged atoms. B.Lone pairs of electrons. C.Ions. D.Bonded pairs o
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2 years ago
When you combine 50.0 mL of 0.100 M AgNO3 with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from
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Answer : The enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

Explanation :

First we have to calculate the moles of AgNO_3 and HCl.

\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole

\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole

Now we have to calculate the moles of AgCl formed.

The balanced chemical reaction will be,

AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)

As, 1 mole of AgNO_3 react with 1 mole of HCl to give 1 mole of AgCl

So, 0.005 mole of AgNO_3 react with 0.005 mole of HCl to give 1 mole of AgCl

The moles of AgCl formed  = 0.005 mole

Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml

Now we have to calculate the mass of solution.

Mass of the solution = Density of the solution × Volume of the solution

Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g

Now we have to calculate the heat.

q=m\times C\Delta T=m\times C \times (T_2-T_1)

where,

q = heat

C = specific heat capacity = 4.18J/g^oC

m = mass = 100 g

T_2 = final temperature = 24.21^oC

T_1 = initial temperature = 23.40^oC

Now put all the given values in the above expression, we get:

q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC

q=338.58J

Now  we have to calculate the enthalpy of the reaction.

\Delta H_{rxn}=\frac{q}{n}

where,

\Delta H_{rxn} = enthalpy of reaction = ?

q = heat of reaction = 338.58 J

n = moles of reaction = 0.005 mole

Now put all the given values in above expression, we get:

\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole

Conversion used : (1 KJ = 1000 J)

Therefore, the enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

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3 years ago
What is the percent yield of ferrous sulfide if the actual yield is 220.0g and the theoretic yield is 275.6?
Amanda [17]
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