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Reika [66]
3 years ago
8

so they are asking me to find the empirical formula for a compound that is 7.70% carbon and 92.3% chlorine. Can you show me step

by step how to solve this step
Chemistry
1 answer:
Nikolay [14]3 years ago
5 0

Answer:

Empirical formula is CCl₄

Explanation:

Given data:

Percentage of carbon = 7.70%

Percentage of chlorine = 92.3%

Empirical formula = ?

Solution:

Number of gram atoms of Cl = 92.3 / 35.5 = 2.6

Number of gram atoms of C = 7.70 / 12 = 0.64

Atomic ratio:

            C                      :              Cl            

           0.64/0.64        :             2.6/0.64

            1                      :                4        

C : Cl  = 1 : 4

Empirical formula is CCl₄.

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Bismuth-210 has a half life of 5.0 days if you had a sample of 60.00g of Bismuth-210 how much would be left after 15 days? (Show
Zielflug [23.3K]

Answer:

7.5 gm left

Explanation:

Bismuth-210 has a half life of 5 days

15 days is 15/5 = 3 half lives

since half the amount is left in 5 days or 1 half life

(1/2) x (1/2) x (1/2) the staring amount would be left in

3 half lives. so 1/8 is left

(1/8) x 60.0 = 7.5 gm left

5 0
3 years ago
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Please don’t listen to those people who always put links those are scams. The answer is the kidney, ureters, bladder, and urethra. This system filters your blood,removing waste and excess water. Your welcome!
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3 years ago
calculate the mass of 120cc nitrogen present at STP. how many number of molecules are present in it?​
Stells [14]

Answer:

0.15008\ \text{g}

3.23\times 10^{21}

Explanation:

1 mol of nitrogen at STP = 22.4 L = 22400 cc

n = Mol of N_2 = \dfrac{120}{22400}=0.00536\ \text{mol}

M = Molar mass of N_2 = 28\ \text{g/mol}

N_A = Avogadro's number = 6.022\times 10^{23}\ \text{mol}^{-1}

Mass of N_2 is

m=nM\\\Rightarrow m=0.00536\times 28\\\Rightarrow m=0.15008\ \text{g}

Mass of the nitrogen is 0.15008\ \text{g}

Number of molecules is given by

nN_A=0.00536\times 6.022\times 10^{23}=3.23\times 10^{21}\ \text{molecules}

The number of molecules present in it are 3.23\times 10^{21}

5 0
3 years ago
Write the symbol for every chemical element that has atomic number greater than 55 and atomic mass less than 144.0 u
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56     Barium       137.328 amu
57     Lanthanium   138.905 amu
58     Cerium       140.116 amu
59     <span>Praseodymium    140.908 amu
60     Neodymium   144.243 amu

Neodymium is already greater than 144 amu. Therefore, these elements only include Barium, Lanthanium, Cerium and Praseodymium.</span>
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Answer:

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