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3 years ago

How many grams of potassium chloride (KCl) can be produced from 180.6 g of chlorine (Cl2)?

Chemistry

1 answer:

GalinKa [24]3 years ago

3 0

Answer:

379.79 g of KCl

Explanation:

The balanced equation for the reaction is given below:

Cl₂ + 2KBr —> Br₂ + 2KCl

Next, we shall determine the mass of Cl₂ that reacted and the mass of KCl produced from the balanced equation. This can be obtained as follow:

Molar mass of Cl₂ = 70.90 g/mol

Mass of Cl₂ from the balanced equation = 1 × 70.90 = 70.90 g

Molar mass of KCl = 74.55 g/mol

Mass of KCl from the balanced equation = 2 × 74.55 = 149.1 g

SUMMARY:

From the balanced equation above,

70.90 g of Cl₂ reacted to produce 149.1 g of KCl.

Finally, we shall determine the mass of KCl produced by the reaction of 180.6 g of Cl₂. This can be obtained as follow:

From the balanced equation above,

70.90 g of Cl₂ reacted to produce 149.1 g of KCl.

Therefore, 180.6 g of Cl₂ will react to produce = (180.6 × 149.1)/70.90 = 379.79 g of KCl.

Thus, 379.79 g of KCl were obtained from the reaction.

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"The elementary reaction 2 NO2(g) → 2 NO(g) + O2(g) is second order in NO2 and the rate constant at 600 K is 6.77 × 10-1 M-1s-1.

blsea [12.9K]

Answer : The half-life at this temperature is, 3.28 s

Explanation :

To calculate the half-life for second order the expression will be:

$t_{1/2}=\frac{1}{k\times [A_o]}$

When,

$t_{1/2}$ = half-life = ?

$[A_o]$ = initial concentration = 0.45 M

k = rate constant = $6.77\times 10^{-1}M^{-1}s^{-1}$

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

$t_{1/2}=\frac{1}{6.77\times 10^{-1}M^{-1}s^{-1}\times 0.45M}$

$t_{1/2}=3.28s$

Therefore, the half-life at this temperature is, 3.28 s

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3 years ago

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saw5 [17]

Explanation:

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2 years ago

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3 years ago

What is the total mass of kno3 that must be dissolved in 50. Grams of h2o at 60.Â°c to make a saturated solution?

Trava [24]

Answer:

55 g

Explanation:

First, we have to look for the solubility of KNO₃ at 60°C, considering that the solubility is the maximum amount of solute that can be dissolved in 100 grams of solvent, that is, the concentration of a saturated solution.

The solubility of KNO₃ at 60°C is 110.0 g of KNO₃ per 100 g of water. The mass of KNO₃ that must be dissolved in 50 g of water to make a saturated solution is:

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6 0

3 years ago

In the laboratory a student determines the specific heat of a metal. She heats 19.6 grams of zinc to 98.37°C and then drops it i

fgiga [73]

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc = 98.37 °C

Initial temperature of water = 24.16 °C

Final temperature of water (and zinc) = 25.70 °C

Specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate Specific heat of zinc

Heat lost by zinc = heat won by water

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

Qzinc = -Qwater

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

7 0

3 years ago