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asambeis [7]
3 years ago
11

How many significant figures are there in 1.01400?

Chemistry
1 answer:
taurus [48]3 years ago
7 0

Answer:

I think it's 4

Explanation:

Because the last two zeros won't be needed

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Which phrase describes the decay modes and the half-lives of K-37 and K-42?
andrew11 [14]

Answer: (3) different decay modes and different half-lives

Explanation: Potassium-37 and Potassium-42 have the same number of protons  but different numbers of neutrons i.e. they have same atomic number but different mass number. They are isotopes. Thus K-37 and K-42 have different decay modes

K-37 is positron decay and K-42 is beta decay.

_{19}^{37}{\textrm {Kr}}\rightarrow _{18}^{37}{\textrm {Ar}}+_{+1}^0{\textrm {e}}

_{19}^{42}{\textrm {Kr}}\rightarrow _{20}^{42}{\textrm {Ca}}+_{-1}^0{\textrm {e}}


K-37 has a half life of 1.23 seconds and K-42 has a half life of 12.4 hrs.

Thus they have different decay modes and different half lives.

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If the pressure of a 2.00 L sample of gas is 50.0 kPa, what pressure does the gas exert if its volume is decreased to 20.0 mL?
dimulka [17.4K]

Answer:

P₂ = 5000 KPa

Explanation:

Given data:

Initial volume = 2.00 L

Initial pressure = 50.0 KPa

Final volume = 20.0 mL (20/1000=0.02 L)

Final pressure = ?

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

50.0 KPa × 2.00L = P₂ × 0.02 L

P₂ = 100 KPa. L/0.02 L

P₂ = 5000 KPa

8 0
3 years ago
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