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Vedmedyk [2.9K]
3 years ago
8

What do gasses have

Chemistry
2 answers:
Goshia [24]3 years ago
6 0

Answer:

Explanation:

Gas is a state of matter that has no fixed shape and no fixed volume. Gases have lower density than other states of matter, such as solids and liquids. There is a great deal of empty space between particles, which have a lot of kinetic energy.

tia_tia [17]3 years ago
4 0

Answer:

Gas is a state of matter that has no fixed shape and no fixed volume. Gases have lower density than other states of matter, such as solids and liquids. There is a great deal of empty space between particles, which have a lot of kinetic energy.

Explanation:

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Can someone help me out please ? I am very stuck here
Alex17521 [72]
Hey there!

<span>Atomic Masses :
</span>
H = <span>1.00794  a.m.u
N = </span><span>14.0067  a.m.u
O = </span><span>15.9994  a.m.u

Therefore:

HNO3 = </span>1.00794 + 14.0067 + ( 15.9994 * 3 ) =>  <span>63.0128 g/mol</span>
7 0
3 years ago
22. Radon has a half-life of 3.83 days. How long will it take a 225 g sample to decay to 14.06 g? (3pts.)
prohojiy [21]

Answer:

15.32 days

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 3.83 days

Original amount (N₀) = 225 g

Amount remaining (N) = 14.06 g

Time (t) =.?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Original amount (N₀) = 225 g

Amount remaining (N) = 14.06 g

Number of half-lives (n) =?

N = N₀ / 2ⁿ

14.06 = 225 / 2ⁿ

Cross multiply

14.06 × 2ⁿ = 225

Divide both side by 14.06

2ⁿ = 225 / 14.06

2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time. This can be obtained as follow:

Half-life (t½) = 3.83 days

Number of half-lives (n) = 4

Time (t) =.?

n = t / t½

4 = t / 3.83

Cross multiply

t = 4 × 3.83

t = 15.32 days

Therefore the time for 225 g sample of Radon to decay to 14.06 g is 15.32 days

8 0
3 years ago
What is the concentration of pb 2 ions in a solution prepared by adding 5. 00 g of lead(ii) iodide to 500. ml of 0. 150 m ki? [
sashaice [31]

37.1×10−8M is the concentration of pb 2 ions in a solution prepared by adding 5. 00 g of lead(ii) iodide to 500. ml of 0. 150 m ki? [ k sp(pbi 2) =1. 4 × 10 –8] by common ion effect.

A phenomenon known as the "common ion effect" allows for the modification of a salt's molar solubility by adding another salt in which one ion dissociates in solution and maintains equilibrium with the undissociated salt.

The term "common-ion effect" describes the reduction in solubility of an ionic precipitate caused by the addition of a soluble molecule that shares an ion with the precipitate to the solution. Le principle for the equilibrium response of ionic association/dissociation leads to this behavior.

To learn more about the "common-ion effect" please visit-
brainly.com/question/13194653
#SPJ4

4 0
2 years ago
Students in Ms. Winters' class were interested in testing the antibacterial strength of hand soap versus hand sanitizer. The stu
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My answer is a please tell me if I helped
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I believe the correct response is the first option.
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