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3 years ago

After cloud droplets form, what must happen to them for precipitation to occur?

1 answer:

5 0

For precipitation to happen, first tiny water droplets must condense on even tinier dust, salt, or smoke particles, which act as a nucleus. Water droplets may grow as a result of additional condensation of water vapor when the particles collide. If enough collisions occur to produce a droplet with a fall velocity which exceeds the cloud updraft speed, then it will fall out of the cloud as precipitation.

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If measurements are all consistent, such as 3.4, 3.6, 3.5, these numbers are said to have

barxatty [35]

A I think I’m sorry I’m not sure

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3 years ago

Why is the mass in amu of a carbon-12 atom reported as 12.011 in the periodic table of the elements?

Bas_tet [7]

The mass of an element listed in the Periodic Table is the weighted average of all its naturally occurring isotopes.

Naturally occurring carbon is about

99 % carbon-12 (12.000 u) + 1 % carbon-13 (13.003 u).

That extra carbon-13 makes the <em>average atomic mass</em> greater than 12.000 u.

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3 years ago

When 108 g of water at a temperature of 23.9 °c is mixed with 66.9 g of water at an unknown temperature, the final temperature o

olganol [36]

Here,

Heat gain by the first sample of water + Heat lost by the second sample of water is equal to zero (0).

Now, Mass of water sample one = 108 g (given)

Mass of water sample two = 66.9 g (given)

Temperature for water sample one = $23.9^{0}C$

Let, temperature for water sample two =x

And, final temperature = $47.2^{0}C$

Now,

$mass of water sample one\times specific heat of water\times (T_{f} - T_{i}) + mass water sample two\times specific heat of water\times (T_{f} - T_{i}) = 0$

where, $T_{f}$ = final temperature

$T_{i}$ = initial temperature

Substitute all the given values in above formula:

$(108 g\times 4.184 J/g . K\times ( 47.2^{0}C- 23.9^{0}C) )+ (66.9 g \times 4.184 J/g . K\times (47.2^{0}C -x))= 0$

$(451.872 J/K \times (23.3^{0}C)) + (279.9096 J/K\times (47.2^{0}C -x)) = 0$

$(10528.6176 J/K(^{0}C)+ (13211.73312 J/K(^{0}C) -279.9096 J/K\times x)= 0$

$(23740.35072 J/K(^{0}C) -279.9096 J/K\times x)= 0$

$-279.9096 J/K\times x= -23740.35072 J/K(^{0}C)$

$x =\frac{23740.35072 J/K(^{0}C)}{279.9096 J/K}$

[tex x =84.81^{0}C [/tex]

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4 years ago

When an atom's outer electron shell is only partially filled, the atom is considered?

kiruha [24]

The Atom it considered Chemically Unstable

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3 years ago

Read 2 more answers

What is the reaction of Dimethylamamine + acetic acid

Arada [10]

DMA is prepared commercially by the reaction of dimethylamine with acetic anhydride or acetic acid. Dehydration of the salt of dimethylamine and acetic acid also furnishes this compound: CH3CO2H·HN(CH3)2 → H2O + CH3CON(CH3) Dimethylacetamide can also be produced by the reaction of dimethylamine with methyl acetate.

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3 years ago