All categories

3 years ago

(02.02 MC)

Chemistry

1 answer:

ratelena [41]3 years ago

7 0

This is due to evaporation

You might be interested in

A balloon at 32 °C is filled with 21 L of air. What would its volume be at a temperature of 52 °C, assuming pressure remains con

I am Lyosha [343]

Answer:C). 24L

Explanation: it works

4 0

4 years ago

To breed or train; to tame a wild animal to live with humans.

Ierofanga [76]

Answer:

The answer is to train.

Explanation:

Training is another word for taming a wild animal. To breed is to take to animals and make them have offspring with each other.

5 0

3 years ago

I NEED THIS ASAP Someone please help me

IceJOKER [234]

Answer:

B. Four moles of water were produced from this reaction.

Explanation:

I took the test and got it correct

8 0

3 years ago

Read 2 more answers

What exacerbates the heat waves?

KIM [24]

Answer:

A heatwave occurs when a system of high atmospheric pressure moves into an area and lasts two or more days.

Explanation:

In such a high-pressure system, air from upper levels of our atmosphere is pulled toward the ground, where it becomes compressed and increases in temperature.

5 0

3 years ago

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as molarity, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

Find the molar mass (M) of potassium hydroxide.

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

Calculate the moles of potassium hydroxide.

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

Convert the volume of water to litres.

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

Calculate the concentration.

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0

3 years ago