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Tju [1.3M]
3 years ago
13

Number of electrons the atom my accept from another atom?

Chemistry
1 answer:
ZanzabumX [31]3 years ago
8 0

Answer:

Yes atoms can accept the electrons from another atom like

Chlorine accept electron from hydrogen.

Explanation:

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20 minus 2 plus 1 equals negative -2000000
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2 years ago
A 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, what is the volume? *
elena-14-01-66 [18.8K]

The new volume when pressure increases to 2,030 kPa is 0.8L

BOYLE'S LAW:

The new volume of a gas can be calculated using Boyle's law equation:

P1V1 = P2V2

Where;

  1. P1 = initial pressure (kPa)
  2. P2 = final pressure (kPa)
  3. V1 = initial volume (L)
  4. V2 = final volume (L)

According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:

406 × 4 = 2030 × V2

1624 = 2030V2

V2 = 1624 ÷ 2030

V2 = 0.8L

Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.

Learn more about Boyle's law calculations at: brainly.com/question/1437490?referrer=searchResults

3 0
2 years ago
What tool does the US government use to keep track of inflation throughout the years?
Alchen [17]
The answer is d the inflation scale
3 0
3 years ago
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What is the average kinetic energy of any substance at 0 K?
Aloiza [94]
The answer is zero!

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5 0
3 years ago
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A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h
frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]

Where;

\frac{f_2}{f_1}  is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1}  = 10^{0.7289}\\\\\frac{f_2}{f_1}  = 5.356

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0
3 years ago
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