1A. A strontium hydroxide solution is prepared by dissolving 10.45 g of Sr(OH)2 in water to make 41.00 mL of solution. What is the molarity of this solution? 1B. Next the strontium hydroxide solution prepared in part (a) is used to titrate a nitric acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between strontium hydroxide and nitric acid solutions.
1C. If 23.9 mL of the strontium hydroxide solution was needed to neutralize a 31.5 mL aliquot of the nitric acid solution, what is the concentration (molarity) of the acid?

A. The molarity of the Sr(OH)₂ solution is 2.09 M

B. The balanced equation for the reaction is

2HNO₃ + Sr(OH)₂ —> Sr(NO₃)₂ + 2H₂O

C. The molarity of the acid, HNO₃ is 3.17 M

A. Determination of the molarity of the Sr(OH)₂ solution

Mass of Sr(OH)₂ = 10.45 g

Molar mass of Sr(OH)₂ = 88 + 2(16 + 1) = 122 g/mol

Mole = mass / molar mass

Mole of Sr(OH)₂ = 10.45 / 122

Mole of Sr(OH)₂ = 0.0857 mole

Volume = 41 mL = 41 / 1000 = 0.041 L

Molarity = mole / Volume

Molarity of Sr(OH)₂ = 0.0857 / 0.041

B. The balanced equation for the reaction.

C. Determination of the molarity of the acid, HNO₃.

From the balanced equation above,

The mole ratio of the acid, HNO₃ (nA) = 2

The mole ratio of the base, Sr(OH)₂ (nB) = 1

From the question given above,

Volume of base, Sr(OH)₂ (Vb) = 23.9 mL

Molarity of base, Sr(OH)₂ (Mb) = 2.09 M

Volume of acid, HNO₃ (Va) = 31.5 mL

MaVa / MbVb = nA/nB

(Ma × 31.5) / (2.09 × 23.9) = 2

(Ma × 31.5) / 49.951 = 2

Cross multiply

Ma × 31.5 = 49.951 × 2

Ma × 31.5 = 99.902

Divide both side by 31.5

Ma = 99.902 / 31.5

Thus, molarity of the acid, HNO₃ is 3.17 M

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1. 0.00352 M

2. 2HNO3(aq) + Sr(OH)2(aq) > Sr(NO3)2(aq) + 2H2O(l)

3. 0.00534 M

Explanation:

1.

Mass of strontium hydroxide= 10.45 g

Volume of solution = 41.00 ml

Number of moles = mass of Sr(OH)2/molar mass of Sr(OH)2 = 10.45g/121.63 g/mol= 0.0859 moles

Molarity= number of moles × volume = 0.0859 ×41/1000 = 0.00352 M

2.

2HNO3(aq) + Sr(OH)2(aq) > Sr(NO3)2(aq) + 2H2O(l)

3.

Concentration of acid CA= the unknown

Volume of acid VA= 31.5 ml

Concentration of base CB= 0.00352 M

Volume of base VB= 23.9 ml

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB = NA/NB

CAVANB= CBVBNA

CA= CBVBNA/VANB

CA= 0.00352 × 23.9 ×2/31.5 ×1

CA= 0.00534 M

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.