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2 years ago

How does increasing temperature affects reaction rate?

Chemistry

1 answer:

valkas [14]2 years ago

4 0

Answer: It increases the reaction rate.

Step-by-step explanation:

The particles gain kinetic energy.

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A sample that contains only SrCO3 and BaCO3 weighs 0.800 g. When it is dissolved in excess acid, 0.211 g car- bon dioxide is lib

MrRissso [65]

Answer:

53.9%

Explanation:

1 mole of BaCO₃ yields 1 mole of CO₂,

1 mole of SrCO₃ yields 1 mole of CO₂

m₁ = mass of BaCO₃

m₂ = mass SrCO₃

molar mass of SrCO₃ = 147.63 g/mol

molar mass of BaCO₃ = 197.34 g/mol

molar mass of CO₂ = 44.01 g/mol

mole of CO₂ in 0.211 g = 0.211 g / 44.01 = 0.00479

mole of BaCO₃ = m₁ / 197.34

mole of SrCO₃ = m₂ / 147.63

mole of BaCO₃ + mole of SrCO₃ = 0.00479

(m₁ / 197.34) + (m₂ / 147.63) = 0.00479

147.63 m₁ + 197.34 m₂ = 139.55

m₁ + m₂ = 0.8

m₁ = 0.8 - m₂

147.63 (0.8 - m₂) + 197.34 m₂ = 139.55

118.104 - 147.63 m₂ + 197.34 m₂ = 139.55

49.71 m₂ = 139.55 - 118.104 = 21.446

m₂ = 21.446 / 49.71 = 0.431

the percentage of m₂ ( SrCO₃ ) in the sample = 0.431 / 0.8 = 0.539 × 100 = 53.9%

5 0

3 years ago

6. Would you describe each of these temperatures as warm, hot, or cold?

mash [69]

Answer:

b: Hot

a: Cold

c: Cold

d: Warm

e: Warm

f: Cold

g: Hot

Explanation:

8 0

3 years ago

Why do boys play with your emotions?

mamaluj [8]

I dont think always kow how to treat a woman right

6 0

3 years ago

Read 2 more answers

Convert 15 meters to kilometers

Gelneren [198K]

Answer:

the answer is 0.015

8 0

3 years ago

Read 2 more answers

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = $1.83g/cm^3=1.83g/ml$

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L$

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg$

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0

3 years ago