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Marianna [84]
3 years ago
11

Find the iupac name for this question​

Chemistry
1 answer:
Vanyuwa [196]3 years ago
8 0

Answer:

3 - Methylpent-2-yn-1-ol

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Question 6 A certain ionic compound has a solubility in water of at . Calculate the greatest mass of that could be dissolved in
emmainna [20.7K]

Complete Question:

A certain ionic compound X has a solubility in water of .765g/ml at 5 degrees C. Calculate the mass of X that's dissolved in 3 L of a saturated solution of X in the water at this temp. Be sure your answer has the correct unit symbol and significant digits.

Answer:

2.29 kg

Explanation:

A saturated solution is a solution that has the maximum amount of solute diluted at it, so if more solute is added, it will precipitate. The solutions can also be unsaturated (when more solute can be dissolved) or supersaturated (when there is more solute dissolved than the maximum. It's a very unstable solution).

The saturation is measured by the solubility, which indicates how much mass of the solute can be added to a certain volume of the solvent. So, the solubility (S) is the mass (m) divided by the volume (V).

For a solution with 3 L = 3000 mL,

S = m/V

0.765 = m/3000

m = 3000 * 0.765

m = 2295 g

m = 2.29 kg

7 0
4 years ago
A reactant decomposes with a half-life of 29.5 s when its initial concentration is 0.229 M. When the initial concentration is 0.
Dmitrij [34]

Answer :

The order of reaction is, 0 (zero order reaction).

The value of rate constant is, 0.00388Ms^{-1}

Explanation :

Half life : It is defined as the time in which the concentration of a reactant is reduced to half of its original value.

The general expression of half-life for nth order is:

t_{1/2}\propto \frac{1}{[A_o]^{n-1}}

or,

\frac{t_{1/2}_1}{t_{1/2}_2}=\frac{[A_2]^{n-1}}{[A_1]^{n-1}}

or,

n=\left(\frac{\log\frac{(t_{1/2})_1}{(t_{1/2})_2}}{\log\frac{(A)_2}{(A)_1}}\right )+1           .............(1)

where,

t_{1/2} = half-life of the reaction

n = order of reaction

[A] = concentration

As we are given:

Initial concentration of A = 0.229 M

Final concentration of A = 0.639 M

Initial half-life of the reaction = 29.5 s

Final half-life of the reaction = 82.3 s

Now put all the given values in the above formula 1, we get:

n=\left (\frac{\log \frac{29.5}{82.3}}{\log\frac{0.639}{0.229}}\right )+1

n=0.000196\approx 0

Thus, the order of reaction is, 0 (zero order reaction).

Now we have to determine the rate constant.

To calculate the rate constant for zero order the expression will be:

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

When,

t_{1/2} = 29.5 s

[A_o] = 0.229 M

29.5s=\frac{0.229M}{2k}

k=0.00388Ms^{-1}

Thus, the value of rate constant is, 0.00388Ms^{-1}

4 0
3 years ago
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