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Zina [86]
2 years ago
13

What is the percentage of oxygen in carbon dioxide (CO2)

Chemistry
1 answer:
pychu [463]2 years ago
5 0

Explanation:

carbon dioxide has a percent composition of 72.7% oxygen, i.e. for every 100 g of carbon dioxide you get 72.7 g of oxygen.

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Name two elements that have properties similar to those of beryllium and have average atomic masses higher than 130
sertanlavr [38]
In the periodic table, elements of the same group are characterized by having the same similar properties.
So, first we will check the elements that lie within the same group as <span>beryllium  and then we will attempt to choose the elements with atomic mass higher than 130.

So, elements in the same group as </span>beryllium are: magnesium, calcium, strontium, barium and radium.
Among these elements, we will find that:
radium has atomic mass of 226 amu
barium has atomic mass of 137.327 amu

Based on this, the two elements would be barium and radium.

3 0
3 years ago
Compute the ionization energy for a single atom of hydrogen
cricket20 [7]
The ionization energy<span> for </span>hydrogen<span> is 1312 kilojoules per mole. This is the same ... Electrically neutral </span>atoms<span> include a </span>single<span> proton and electron held together.</span>
6 0
3 years ago
How many electrons does Br lose or gain to fill its octet?
Greeley [361]

i think it is 8. I might be wrong.

4 0
3 years ago
A chemist wishing to do an experiment requiring 47-Ca2+ (half-life = 4.5 days) needs 5.0 μg of the nuclide. What mass of 47-CaCO
NARA [144]

Answer:

5.8μg

Explanation:

According to the rate or decay law:

N/N₀ = exp(-λt)------------------------------- (1)

Where N = Current quantity,  μg

            N₀ = Original quantity, μg

             λ= Decay constant day⁻¹

              t =  time in days

Since the half life is 4.5 days, we can calculate the  λ from (1) by  substituting N/N₀ = 0.5

0.5 = exp (-4.5λ)

ln 0.5  = -4.5λ

-0.6931 = -4.5λ

λ =   -0.6931 /-4.5

  =0.1540 day⁻¹

Substituting into (1)  we have :

N/N₀ = exp(-0.154t)----------------------------- (2)

To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:

N = 5.0 μg

N₀ = Unknown

t = 1 day

Substituting into (2) we have

[5/N₀]   = exp (-0.154 x 1)

    5/N₀        = 0.8572

N₀  =  5/0.8572

     =    5.8329μg

    ≈     5.8μg

The Chemist must order 5.8μg  of 47-CaCO3

6 0
2 years ago
Which of the following is not a way populations are described?​
gladu [14]
The last answer

the place where the organisms live

Hope this helps!!! :))))
4 0
2 years ago
Read 2 more answers
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