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skelet666 [1.2K]
2 years ago
14

Large air masses moving across and area are called _____?

Chemistry
2 answers:
faltersainse [42]2 years ago
6 0
Fronts. This includes cold fronts and warm fronts. Hoped this helped!
docker41 [41]2 years ago
6 0
Answer: Air masses





Explanation: An air mass is a large body of air that has about the same conditions throughout. For example, an air mass might have cold dry air. Another air mass might have warm moist air. The characteristics of an air mass depend on where the air mass formed. The air must stay over that location long enough to pick up those characteristics.
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A chemist adds of a M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has add
weqwewe [10]

corrected question: A chemist adds 135mL of a 0.21M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.

Answer:

5.37g

Explanation:

0.21M means ; 0.21mol/dm³

1dm³=1L , so we can say 0.21mol/L

if 0.21mol of Zinc nitrate is contained in 1L of water

x   will be contained in 135mL of water

x= 0.21*135*10³/1

=0.02835moles

number of moles=  mass/ molar mass

mass= number of moles *molar mas

molar mass of Zn(NO₃)₂=189.36 g/mol

mass= 0.02835 *189.36

mass=5.37g

3 0
3 years ago
Melting can be best described as a process in which molecules
Savatey [412]
Gain enough kinetic energy to get past each other. Ad you heat up a substance, the temperature increases as does the kinetic energy of the particles. At a point the temperature of the substance will stop increasing. The energy is now being used to increase the potential and move the particles further apart.
5 0
3 years ago
Read 2 more answers
Consider the following reaction: 2{\rm{ N}}_2 {\rm{O(}}g)\; \rightarrow \;2{\rm{ N}}_2 (g)\; + \;{\rm{O}}_2 (g)
fredd [130]

Answer:

a. 5.9 × 10⁻³ M/s

b. 0.012 M/s

Explanation:

Let's consider the following reaction.

2 N₂O(g) → 2 N₂(g) + O₂(g)

a.

Time (t): 12.0 s

Δn(O₂): 1.7 × 10⁻² mol

Volume (V): 0.240 L

We can find the average rate of the reaction over this time interval using the following expression.

r = Δn(O₂) / V × t

r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s

r = 5.9 × 10⁻³ M/s

b. The molar ratio of N₂O to O₂ is 2:1. The rate of change of N₂O is:

5.9 × 10⁻³ mol O₂/L.s × (2 mol N₂O/1 mol O₂) = 0.012 M/s

4 0
3 years ago
How many moles are in 3.5 x 10^24 molecules of methane gas?
Alexxx [7]

Answer:

5.81 moles

Explanation:

To find the number of moles (n) in 3.5 x 10²⁴ molecules of methane gas, we divide the number of molecules by Avagadro's number (nA). That is,

n = number of molecules ÷ 6.02 × 10²³

According to this question, 3.5 x 10^24 molecules of methane gas was given, hence,

n = 3.5 × 10²⁴ ÷ 6.02 × 10²³

n = 3.5/6.02 × 10(24 - 23)

n = 0.5814 × 10¹

n = 5.81 moles

7 0
3 years ago
HELP NOW Do Now 3.5 Purple
Serga [27]

Answer:

The smell of a chocolate is from the presence of volatile compounds present in the chocolate bar which at room temperature readily changes phase from solid to liquid to vapor or gas

Explanation:

There are nearly 600 identified compounds present in a chocolate bar and out of these, there are volatile components which gives the chocolate bar its distinctive aroma.

These volatile chocolate contents readily change phase from solid to vapor, with very short duration liquid phase.

For example, 3 methylbutanal, vanillin, and several organic compounds which are known to be readily volatile.

4 0
2 years ago
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