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3 years ago

How does the type of precipitation change currently?

Chemistry

1 answer:

Inga [223]3 years ago

5 0

Answer:

There are a couple different types of precipitation, though it is generally taken to mean a synonym for rain. It includes rain, snow, sleet, and hail. With the increase in global warming, there is more precipitation. In general, the type of precipitation varies with the season; we see less in the summer of all types, more snow, sleet, and hail in the winter, and more rain in the temperate seasons like fall.

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What is the mass of 5 mole of ammonia . Calculate the number of NH₃ molecules, nitrogen atom and hydrogen atoms in it..

Aloiza [94]

Molar mass of NH_3

$\\ \sf\longmapsto 14u+3(1u)$

$\\ \sf\longmapsto 14u+3u$

$\\ \sf\longmapsto 17g/mol$

We know.

No of moles=Given mass/Molar mass

$\\ \sf\longmapsto Given\;Mass=17(5)$

$\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g$

Now

Lets write the balanced equation

$\\ \sf\longmapsto N_2+3H_2=2NH_3$

There is 2moles of Ammonia
3moles of H_2
1mole of N_2

Now

$\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}$

For Hydrogen

$\\ \sf\longmapsto 3\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 18.069\times 10^{23}$

$\\ \sf\longmapsto 1.8\times 10^{22}molecules$

For Ammonia

$\\ \sf\longmapsto 2\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 12.046\times 10^{23}$

$\\ \sf\longmapsto 1.2\times 10^{22}molecules$

For Nitrogen

$\\ \sf\longmapsto 1\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 6.023\times 10^{23}molecules$

3 0

2 years ago

Water molecules are highly _______ and are always _________.

anyanavicka [17]

Water molecules are highly packed and are always near each other.

7 0

2 years ago

Explain hydrogen bonding.

denis23 [38]

Answer:

Hydrogen bonding, interaction involving a hydrogen atom located between a pair of other atoms having a high affinity for electrons; such a bond is weaker than an ionic bond or covalent bond but stronger than van der Waals forces. Hydrogen bonds can exist between atoms in different molecules or in parts of the same molecule.

Explanation:

3 0

3 years ago

When a mixture of 12.0 g of acetylene (c2h2 and 12.0 g of oxygen (o2 is ignited, the resultant combustion reaction produces co2

valina [46]

The product of the complete combustion of any fuel (in this case, acetylene) are indeed water and carbon dioxide.
Balancing the combustion reaction,
C2H2 +(5/2) O2 --> 2CO2 + H2O
The number of moles of C2H2 will be,
(12 g) x (1 mole/26 g) = 6/13 mole
Then, the number of moles of O2 is,
(12 g) x (1 mole/32 g) = 3/8 mole
Therefore the limiting reaction is the O2. Getting the amount of CO2 and H2O produced from balancing,
CO2 = (3/8 moles) x (2 moles CO2/ 5/2 mole O2)(44 g/ 1 mole) = 52.8 g
H2O = (3/8 moles) x (1 mole / 5/2 mole O2)(18 g / 1 mole) = 2.7 g

3 0

3 years ago

Read 2 more answers

Consider the diagram below

Artyom0805 [142]

Energy of the reactants would be correct

5 0

2 years ago