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4 years ago

Which of the following summertime activities consists solely of physical change?

1 answer:

5 0

A: making s sandcastle. This is because water and sand is only a mixture, so they do not react with each other. All the rest include chemical reactions!

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(a) a 0.2 m potassium hydroxide solution is titrated with a 0.1 m nitric acid solution. (i) balanced equation: (ii)what would be

Stells [14]

$\text{KOH} (aq) + \text{HNO}_3 (aq) \to \text{KNO}_3 (aq) + \text{H}_2\text{O} (l)$

The solution shall contain only $\text{KNO}_3 (aq)$ (and water) at the equivalence point. Both potassium hydroxide and nitric acid exist as strong electrolytes. As a result, $\text{KNO}_3 (aq)$ , the salt derived from a reaction between the two species would undergo hydrolysis of a negligible extent. This neutralization reaction therefore be neutral at the equilibrium point.

The question states that the solution is "titrated with a ... nitric acid solution" indicating that $\text{HNO}_3$ is added to the initially-basic solution. PH value of the solution would keep decreasing as the volume of the acid added increases. The final solution would be acidic as it contains not only water and $\text{KNO}_3 (aq)$ , but some $\text{HNO}_3$ as well. Bromothymol blue would therefore demonstrates a yellow color, the color it present in an acidic solution, at the end of the titration.

5 0

3 years ago

A sample of hydrogen gas collected over water occupied 30.0 mL at 24 °C on a day when the atmospheric pressure was 736 Torr. Wha

givi [52]

Answer:0.026ml

Explanation:

Details are found in the image attached. We must subtract the saturated vapour pressure of hydrogen gas at the given temperature from the total pressure of the hydrogen gas collected over water to obtain the actual pressure of hydrogen gas and substitute the value obtained into the general gas equation. The dry hydrogen gas has no saturated vapour pressure hence the value is substituted as given. All temperatures must be converted to Kelvin before substitution.

4 0

4 years ago

Two basic properties of the liquid phase

melamori03 [73]

Fixed density
Particles move smoothly

6 0

4 years ago

PLEASE HELPP<br> Will mark brainliest

zheka24 [161]

Answer:

Explanation:

7 0

3 years ago

Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta

scoray [572]

Answer:

70.77 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as rate of change of volume with respect to time.

Rate of Effusion= $\frac{Volume}{Time}$

Effusion rate of oxygen gas after time t = $E=\frac{4.64 mL}{t}$

Molar mass of oxygen gas = M = 32 g/mol

Effusion rate of unknown gas after time t = $E'=\frac{3.12 mL}{t}$

Molar mass of unknown gas = M'

The rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{E}{E'}=\sqrt{\frac{M'}{M}}$

$\frac{\frac{4.64 mL}{t}}{\frac{3.12 mL}{t}}=\sqrt{\frac{M'}{32 g/mol}}$

M' = 70.77 g/mol

70.77 g/mol is the molar mass of the unknown gas.

8 0

3 years ago