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4 years ago

Which air pressure system is generally associated with pleasant, clear weather?

Chemistry

2 answers:

Anton [14]4 years ago

5 0

D, High Air Pressure. :)

vagabundo [1.1K]4 years ago

3 0

Clear weather is often associated with high air pressure. An area of low pressure would mean that precipitation and clouds would be near.

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The electrolyte that is used to make Hypochlorous Acid is made from Sodium Chloride (NaCl). the bottle says that the electrolyte

kumpel [21]

Answer:

63.9 grams. Yes, the Nacl was converted. Maximum possible ppm is 540ppm.

Explanation:

I this is college level chemistry not regular high school chem.

7 0

3 years ago

PLS HELP The average atomic mass of carbon is 12.01 amu. Based on the atomic

Masteriza [31]

Answer:

B. There is a very large percentage of C-12.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize that, since the average atomic mass is 12.01 amu, then the C-12, with an atomic mass of 12.000 am prevails over C-13 with an atomic mass of 13.003 amu as long as the average is nearer to the former.

In such a way, the answer will be B. There is a very large percentage of C-12.

Regards!

7 0

3 years ago

If the volume of a cylinder is reduced from 8.0 liters to 4.0 liters, the pressure of the gas in the cylinder will change from 7

Elza [17]

Answer:

P₂ = 140 KPa

Explanation:

Given data:

Initial volume = 8.0 L

Final volume = 4.0 L

Initial pressure = 70 KPa

Final pressure = ?

Solution:

According to Boyle's law

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

P₂ = 70 KPa ×8.0 L/4.0 L

P₂ = 560 KPa .L / 4.0 L

P₂ = 140 KPa

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3 years ago

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ira [324]

Delete the cactus and the tree on the right for the first one and the third picture delete the lion moose Buffalo

8 0

3 years ago

Read 2 more answers

Element X has two isotopes. If 72.0% of the element has an isotope mass of 84.9 atomic mass units, and 28.0% of the element has

bija089 [108]

<h2>Answer:</h2>

Average atomic mass of an element is the sum of the masses of its isotopes each multiplied by its natural abundance

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ \sum\limits \: \% age \: of \: each \: isotope \times Atomic \: mass }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 72 \times84.9 + 28 \times 87 }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 6112.8 + 2436 }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 8548.8 }{100} \\$

$\footnotesize \longrightarrow \: \bf Average \: atomic \: mass = 85.488 \: amu \\$

8 0

3 years ago