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Archy [21]
3 years ago
7

Oxygen 22 has a half life of 2.25 seconds. What percent of the sample is left after 2.25 seconds?

Chemistry
2 answers:
Nonamiya [84]3 years ago
6 0

Answer:

50 because it’s a ratio problem

Explanation:

serg [7]3 years ago
3 0

Answer:

So we got 50% left after 2.25 seconds

Explanation:

The left sample is given by

m=m_{0}.2^\frac{-t}{T}

where T is the half-life

After 2.25 s

m=m_{0}.2^\frac{-2.25}{2.25}=m_{0}.\frac{1}{2}

You might be interested in
You have 0.21 moles of Al how many atoms do you have
ZanzabumX [31]

Number of atoms : 1.26 x 10²³

<h3>Further explanation  </h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance  

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

0.21 moles of Al, so n = 0.21

Number of atoms :

\tt N=0.21\times 6.02\times 10^{23}\\\\N=1.26\times 10^{23}

8 0
2 years ago
A sample of pure sodium-24 has an initial mass of 10.0 grams. How many grams of sodium-24 will remain at the end of 4 half-lives
astra-53 [7]
I got b 5)A sample of pure sodium-24 has an initial mass of 10.0 grams. How many grams of sodium-24 will remain at the end of 4 half-lives?


A. 0.315
B. 0.630
C. 5.00
D. 6.00
8 0
3 years ago
Radon-222 is a radioactive gas with a half-life of 3.82 days. How long would it take for fifteen-sixteenths of a sample of radon
Fittoniya [83]
Use the formula in terms of half life from the normal exponential functions 
<span>
N(t) = N(0) (1/2) ^ (t/thalf) </span>
<span>
N(0) is the original quantity </span>
<span>
N(t) = quantity remaining at time t </span>
<span>
t is the time </span>
<span>

thalf is half life </span>
<span>
1/16 = (1/2)^(t/3.82) </span>
<span>
16 = 2^(t/3.82) </span>
<span>
4 = t/3.82 </span>
<span>
t = 15.28 days

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
8 0
3 years ago
Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --&gt; 2CO2(g) + 2 H2O
Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.5kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.8kJ

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.5kJ)=-787kJ

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.8kJ)=-571.6kJ

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)

\Delta H=52.4kJ

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

7 0
3 years ago
30 treadmills to 36 elliptical machines
scoundrel [369]

Answer:

what? that's 66 total, 6 more elliptical machines, a 1 to 1.2 ratio

but I don't know what else you would mean

7 0
2 years ago
Read 2 more answers
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