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wariber [46]
3 years ago
8

15 points

Chemistry
1 answer:
Usimov [2.4K]3 years ago
8 0
A new species of plant develops and another species of plant dies out
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Is H-3 stable?<br> Is He-3 stable?<br> Is He-5 stable?<br> Is Li-7 stable?<br> Is Be-7 stable?
kiruha [24]

Answer:

no, yes, no, yes, yes

Explanation:

good luck!

8 0
3 years ago
How many moles of water would be produced from 3 moles of oxygen in the following reaction? Don’t forget to include units in you
wariber [46]

Answer:

6moles of water

Explanation:

Given parameters:

Number of moles of oxygen  = 3moles

  Reaction equation:

      2H₂   +   O₂   →   2H₂O

Unknown:

Number of moles of water formed  = ?

Solution:

To solve this problem;

      compare the number of moles in the reaction.

        1 mole of oxygen gas will produce 2 mole of water

        3 moles of oxygen gas will produce 3 x 2  = 6moles of water.

7 0
3 years ago
At what temperature would 0.500 moles of gas particles stored in a 100.0 mL container reach a pressure of 15.0 atm?
MArishka [77]
Hey there!

<span>In this case, the equation of Clapeyron is used :

R = 0.082 

Volume in liters :

100.0 mL / 1000 => 0.1 L 
</span>
P * V = n * R * T

15.0 * 0.1 = 0.500 * 0.082 * T

1.5 = 0.041 * T

T = 1.5 / 0.041

T = 36.5 K

Answer C
6 0
3 years ago
Read 2 more answers
The half-life of radium-226 is 1590 years. if a sample contains 100 mg, how many mg will remain after 1000 years?
AlladinOne [14]

Answer:

64.52 mg.

Explanation:

The following data were obtained from the question:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Final amount (N) =.?

Next, we shall determine the rate constant (K).

This is illustrated below:

Half life (t½) = 1590 years

Rate/decay constant (K) =?

K = 0.693 / t½

K = 0.693/1590

K = 4.36×10¯⁴ / year.

Finally, we shall determine the amount that will remain after 1000 years as follow:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Rate constant = 4.36×10¯⁴ / year.

Final amount (N) =.?

Log (N₀/N) = kt/2.3

Log (100/N) = 4.36×10¯⁴ × 1000/2.3

Log (100/N) = 0.436/2.3

Log (100/N) = 0.1896

Take the antilog

100/N = antilog (0.1896)

100/N = 1.55

Cross multiply

N x 1.55 = 100

Divide both side by 1.55

N = 100/1.55

N = 64.52 mg

Therefore, the amount that remained after 1000 years is 64.52 mg

7 0
4 years ago
The kilogram is the standard unit of
Sati [7]

Answer:

The kilogram is the standard unit of mass.

7 0
3 years ago
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