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3 years ago

Why is a high temperature needed to melt potassium iodide?

Chemistry

2 answers:

castortr0y [4]3 years ago

6 0

Explanation:

Potassium Iodide -

It is a salt , which is made of the ions of potassium and iodine ,

Potassium being the cation , and iodide being the anion .

The ions of this salt are held by strong ionic forces , which hold the ions together very strongly .

Hence , a lot of energy is required to melt the salt , as the bond between the ions is very strong .

Hence , the melting point of this salt , i.e. , Potassium Iodide , is very strong .

goldfiish [28.3K]3 years ago

4 0

Potassium Iodide have a lot of strong bonds which requires a very high temperature (high energy) to break them and change into a different state of matter.

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What is the epicenter?

hoa [83]

Answer:

basically the middle of an earthquake

5 0

3 years ago

The reaction A + 2B → products was found to follow the rate law: rate = k[A] 2[B]. Predict by what factor the rate of reaction w

Alexus [3.1K]

Answer:

By a factor of 12

Explanation:

For the reaction;

A + 2B → products

The rate law is;

rate = k[A]²[B]

As you can see, the rate is proportional to the square of the concentration of A and the of the concentration of B .

Let's say initially, [A] = x, [B] = y

The rate law in this case is equal to;

rate1 = k. x².y

Now you double the concentration of A and triple the concentration of B.

[A] = 2x, [B] = 3y

The new rate law is given as;

rate2 = k . (2x)². (3y)

rate2 = k . 4x² . 3y

rate2 = 12 k . x² . y

Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12

Therefore the rate has increased by a factor of 12.

5 0

3 years ago

Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.

weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

7 0

3 years ago

Calculate ΔGrxn for this equation, rounding your answer to the nearest whole number.

Dovator [93]

Answer:

-958 spontaneuos

Explanation:

7 0

3 years ago

Read 2 more answers

Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.

lianna [129]

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i) 0.115M 0 0

ΔC -x +x +x

C(eq) 0.115M - x x x

≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

4 0

3 years ago