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xxTIMURxx [149]
3 years ago
14

What is standard pressure in atmospheres? a.10 b.1 c.5 d.2

Chemistry
1 answer:
lesya [120]3 years ago
6 0
The answer is d. 2 hope this helps
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How can you find the number of neutrons that are in an atom of an element
Temka [501]

Answer:

You just have to take the periodic table of the elements, take the element that interests you. Then, you look at the formula of the element, at the top left you find the number of neutrons and protons and at the bottom left you find the number of protons. Then you just have to make the number from the top left minus the number from the bottom left.

Explanation:

4 0
3 years ago
How many moles are there in 7.4 X 1023 molecules of AgNO3?
vichka [17]
There are 209 g of moles
8 0
3 years ago
555mL of gas is combined at standard temperature and pressure(STP). The volume changes to 660.mL at 30.5 °C. What is the new pre
liberstina [14]

Answer:0.94

Explanation:

4 0
3 years ago
what mass of carbon dioxide gas would be produced if 10g of calcium carbonate reacted with an excess of hydrochloric acid?
LuckyWell [14K]
Answer is: 4,4 grams <span>of carbon dioxide gas would be produced.
</span>Chemical reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
m(CaCO₃) = 10 g.
n(CaCO₃) = 10 g ÷ 100 g/mol.
n(CaCO₃) = 0,1 mol.
From chemical reaction: n(CaCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0,1 mol.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) = 0,1 mol· 44 g/mol.
m(CO₂) = 4,4 g.
7 0
3 years ago
131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___
inn [45]
For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,         
8.04 days    = ln 2 / λ                         
λ   = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at  t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ   = ln 2 / 8.04 days = 0.693 / 8.04 days
t    = 13.0 days 

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

8 0
3 years ago
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