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Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)

marin [14]

Answer:

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

$S.G=\frac{D}{d_w}$

$d_w$ = density of water = 1 g/mL

$D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL$

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

Volume = $\frac{Mass}{Density}$

$=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L$

1 mL = 0.001 L

$Molarity = \frac{n}{V(L)}$

n = number of moles of compound

V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

$Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L$

5 0

3 years ago

The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of th

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Tomato potato A is the answer fo sho

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the specific temperature at which a substance changes from a liquid to a gas is an example of a a. chemical bond. b. chemical ch

9966 [12]

Physical property because going from a liquid to a gas doesn't change it's chemical composition only it's physical state

4 0

3 years ago

Read 2 more answers

In what type of solution was the human cell placed

White raven [17]

Answer:

Isotonic Solution.

Explanation:

There are three types of solutions that can occur in your body based on solute concentration: isotonic, hypotonic, and hypertonic. An isotonic solution is one in which the concentration of solutes is the same both inside and outside of the cell.

4 0

4 years ago

The rate at which a certain Australian tree cricket chirps is 194/min at 28°C, but only 47.6/min at 5°C, From these data calcula

uysha [10]

Answer: The energy of activation for the chirping process is 283.911 kJ/mol

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_2$ = rate of reaction at $28^0C$ = 194/min

$K_1$ = rate of reaction at $5^0C$ = 47.6 /min

$Ea$ = activation energy

R = gas constant = 8.314 J/Kmol

tex]T_1[/tex] = initial temperature = $5^oC=273+5=278K$

tex]T_1[/tex] = final temperature = $28^oC=273+28=301K$

Now put all the given values in this formula, we get

$\frac{194}{47.6}=\frac{E_a}{2.303\times 8.314}[\frac{1}{278}-\frac{1}{301}]$

${E_a}=283911J/mol=283.911kJ/mol$

Thus the energy of activation for the chirping process is 283.911 kJ/mol

8 0

4 years ago