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3 years ago

15

What conditions are important in determining if it will rain or snow?

2 answers:

zhannawk [14.2K]3 years ago

7 0

For snow to fall to the ground, the temperature must be cold both up in the clouds where snowflakes form, and down at ground level. If the air near ground level is too warm, the snow will melt on its way down, changing to rain or freezing rain. Moisture is needed to form clouds and precipitation.

monitta3 years ago

6 0

Well for snow to fall to the ground, the temperature has to be 32 degrees. And it has to be cold in the clouds where snowflakes form and fall. If the air at ground level is warm or above 32 degrees, the snow will melt half way down to the ground and the snow will just change to rain.

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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 <u>molarity</u>, 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)

<u>Find the molar mass (M) of potassium hydroxide.</u>

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

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

<u>Calculate the moles of potassium hydroxide.</u>

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

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

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

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

$V = 0.5000L$

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

<u>Calculate the concentration.</u>

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

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

$c = 0.5188(3) \frac{mol}{L}$ <= 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

The pre-exponential constant and activation energy for the diffusion of iron in cobalt are 1.7 x 10-5 m2/s and 273,300 J/mol, re

Kitty [74]

<u>Answer:</u> The temperature of the system will be 1622 K

<u>Explanation:</u>

The equation relating the pre-exponential factor and activation energy follows:

$\log D=\log D_o-\frac{E_a}{2.303RT}$

where,

D = diffusion coefficient = $2.7\times 10^{-14}m^2/s$

$D_o$ = pre-exponential constant = $1.7\times 10^{-5}m^2/s$

$E_a$ = activation energy of iron in cobalt = 273,300 J/mol

R = Gas constant = 8.314 J/mol.K

T = temperature = ?

Putting values in above equation, we get:

$\log (2.7\times 10^{-14})=\log (1.7\times 10^{-5})-\frac{273,300}{2.303\times 8.314\times T}\\\\T=1622K$

Hence, the temperature of the system will be 1622 K

8 0

3 years ago

If you dilute 250 mL of 6.4 M lithium acetate solution to a volume of 750 mL, what will the concentration of this solution be?

Tresset [83]

The answer would be 1006.4

3 0

3 years ago

_______is the combining capacity of an atom of an element

laiz [17]

Answer:

valancy

Explanation:

3 0

3 years ago

Substances which are easily vaporized are called (organic, inorganic, volatile).

Alla [95]

I believe the answer is volatile!

3 0

3 years ago

Read 2 more answers