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mixas84 [53]
3 years ago
9

What happens to that atom of magnesium-24 if it GAINS an ELECTRON?

Chemistry
1 answer:
Phoenix [80]3 years ago
8 0
Billie Eillish’s documentary titled “The World Is A Little Blurry” is out and is streaming on Apple TV :)
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A 32.4 L gas sample at STP is compressed to a volume of 28.4 L, and the temperature is increased to 352 K. What is the new press
Sindrei [870]

Answer:

1.47 atm

Explanation:

Step 1: Given data

  • Initial volume (V₁): 32.4 L
  • Initial pressure (P₁): 1 atm (standard pressure)
  • Initial temperature (T₁): 273 K (standard temperature)
  • Final volume (V₂): 28.4 L
  • Final pressure (P₂): ?
  • Final temperature (T₂): 352 K

Step 2: Calculate the final pressure of the gas

We can calculate the final pressure of the gas using the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

P₂ = P₁ × V₁ × T₂ / T₁ × V₂

P₂ = 1 atm × 32.4 L × 352 K / 273 K × 28.4 L = 1.47 atm

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3 years ago
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2 years ago
Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How m
nlexa [21]

Answer:

The work done and heat absorbed are both -8,1 kJ

Explanation:

The work done in an isobaric process is defined as:

W = -P (Vf - Vi)

Where P is pressure ( 10 atm)

Vf = 10 L

Vi = 2 L

Thus, <em>W = -80 atm×L ≡ -8,1 kJ</em>

This is the work done in expansion of the gas.  As the gas remains at the same temperature, there is no change in internal energy doing that all work was absorbed as heat.

I hope it helps!

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3 years ago
ASAP!
alisha [4.7K]

Today scientist don't believe there's life in this solar system but there is proof of living things that once used to live on mars.  there might be life in our universe and other galaxies.

Explanation:

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3 years ago
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What volume of concentrated (10.2 M) HCl would be required to prepare 1.11 x 104 mL of 1.5 M HC1? Enter your answer in scientifi
Tomtit [17]

Answer:

The required volume is 1.6 x 10³mL.

Explanation:

When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:

C₁ . V₁ = C₂ . V₂

where,

C₁ and V₁ are the concentration and volume of the concentrated solution

C₂ and V₂ are the concentration and volume of the dilute solution

In this case, we want to find out V₁:

C₁ . V₁ = C₂ . V₂

V_{1} = \frac{C_{2}.V_{2}}{C_{1}} = \frac{1.5M \times1.11.10^{4}mL }{10.2M} =1.6\times10^{3} mL

3 0
3 years ago
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