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3 years ago

5. Calculate the decrease in temperature when 6.00 L at 20.0 °C is compressed to 4.00 L.

Chemistry

1 answer:

Kazeer [188]3 years ago

5 0

Answer: 195.3 K

Explanation: Use Charles Law to get T2

Charles Law V1 / T1 = V2 / T2

Derive the equation to find T2

T2 = V2T1 / V1

Convert temperature into K

T1 = 20.0 °C + 273 = 293 K

Substitute the values:

4.00 L x 293 K / 6.00 L = 195. 3 K

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3 years ago

You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution

vfiekz [6]

Answer:

450. g of 0.173 % KCN solution contains 779 mg of KCN.

Explanation:

Mass of the solution = m

Mass of the KCN in solution = 779 mg

Mass by mass percentage of KCN solution = 0.173%

$(m/m)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$0.173\%=\frac{779 mg}{m}\times 100$

$m=\frac{779 mg}{0.173}\times 100= 450,289 mg$

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3 years ago

Describe two ways in which humans have negatively affected groundwater.

ludmilkaskok [199]

Answer: Stressors

Explanation:

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3 years ago

What temperature will the water reach when 10.1 g CaO is dropped into a coffee cup containing 157 g H2O at 18.0°C if the followi

Zepler [3.9K]

Answer:

Final temperature attained by water = 34.6°C

Explanation:

The reaction of CaO and H₂O is an exothermic reaction. The equation of reaction is given below:

CaO + H₂O ----> Ca(OH)₂

The quantity of heat given off, ΔH°rxn = 64.8KJ/mol = 64800J/mol

Number of moles of CaO = mass/molar mass, where molar mass of Ca0 = 56g/mol, mass of CaO = 10.1g

Number of moles of CaO = 10.1g/56g/mol =0.179moles

Quantity of heat given off by 0.179 moles = 64800 *0.179 = 11599.2J/mol

Using the formula, Quantity of heat, q = mass * specific heat capacity * temperature rise.

mass of mixture = (10.1 + 157)g = 167.1g, Initial temperature = 18.0°C

Final temperature(T₂) - Initial temperature(T₁) = Temperature rise

11599.2J/mol = 167.1g * 4.18J/g·°C * ( T₂ - 18.0°C)

11599.2 = 698.478T₂ - 12572.604

11599.2 + 12572.604 = 698.478T₂

698.478T₂ = 24171.804

T₂ = 34.6°C

Therefore, final temperature attained by water = 34.6°C

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3 years ago

If the decomposition of a sample of k c l o 3 kclox3 produces 3. 29 g of o 2 ox2 , what was the mass (g) of the original sample?

____ [38]

The mass (g) of the original sample after decomposition is 8.3983 g.

A decomposition reaction can be described as a chemical reaction wherein one reactant breaks down into or extra merchandise.

explanation:

Reaction 2KClO₃ ⇒ 2KCl + 3O₂

moles 2 2 3

molar mass 122.55 74.55 32

Given, Mass of O₂ = 3.29g ⇒ moles of O₂

= (3.29/32) = 0.1028

3 moles of O₂ produced by 2 moles of KClO₃

Therefore, 0.1028 moles of O₂ produced by (2*0.1028/3) = 0.06853 moles of Kclo₃

Mass of KClo₃ in original sample is = moles * molar mass

= 0.06853 * 122.55

= 8.3983 g

A decomposition response occurs whilst one reactant breaks down into or extra merchandise. this may be represented through the general equation: XY → X+ Y. Examples of decomposition reactions consist of the breakdown of hydrogen peroxide to water and oxygen, and the breakdown of water to hydrogen and oxygen.

Learn more about decomposition here:-brainly.com/question/27300160

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2 years ago