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cluponka [151]
3 years ago
11

A compound that is composed of only carbon and hydrogen contains 85.7% c and 14.3% h by mass. what is the empirical formula of t

he compound?
Chemistry
1 answer:
Anna35 [415]3 years ago
8 0
The empirical formula is CH2
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What is the electron configuration of an element with atomic number 20? A. 1s2 2s2 2p6 3s2 3p5 B. 1s2 2s2 2p6 3s2 3p6 C. 1s2 2s2
Cerrena [4.2K]

D.  The number of electrons equals the atomic number for a neutral element.  Each number after the letter refers to the number of electrons in that shell.  So for D, 2+2+6+2+6+2 = 20 electrons, which is equal to the atomic number.

8 0
3 years ago
Read 2 more answers
(a) if a sample containing 2.00 ml of nitroglycerin is detonated, how many total moles of gas are produced? (b) if each mole of
gregori [183]
Detonation of nitroglycerin: 

4C_3H_5N_3O_9  ------\ \textgreater \  12CO_2+6N_2+O_2+H_2O

<span>Mass nitroglycerin = 2.00 mL x 1.592 g/mL = 3.184 g 
</span>
Moles = mass / molar mass = 3.184<span> g/ 227.0872 g/mol = 0.01402
</span>
the ratio between nitroglycerin and Carbon dioxide is 4 : 12 
So, moles CO2 = 0.01402 x 12 / 4 =0.0420

the ratio between nitroglycerin and N2 is 4 : 6 
moles N2 = 0.01402 x 6 / 4 =0.0841

<span>the ratio between nitroglycerin and O2 is 4 : 1 </span>
moles O2 = 0.01402 x 1 / 4 = 0.0035

<span>the ratio between nitroglycerin and water is 4 : 1 </span>
<span>in the same way moles water = 0.005258 </span>

total moles = 0.0420 + 0.0841 + 0.0035 + 0.005258 = 0.130758

0.130758<span> x 55 = 5.78 L </span>

Mass N2 = 0.0841 mol x 28.0134 g/mol = 2.3548 g


6 0
3 years ago
For a laboratory investigation some students put a strip of shiny metal into a beaker of blue solution and then stored the beake
KatRina [158]

Answer:

b

Explanation:

8 0
3 years ago
Match the following numbers with the correct number of significant digits they possess. Answers may repeat. (4 points)
mezya [45]

(B. 3) 172 All nonzero digits are significant.

(A. 4) 450.0 x 10^3 Trailing zeroes after the decimal point are significant.

(A. 4) 3427 All nonzero digits are significant.

(B. 3) 0.0000455 Leading zeroes are not significant.

(B. 3) 0.00456 Leading zeroes are not significant.

(C. 5) 2205.2 Zeroes between nonzero digits are significant.

(C. 5) 107.20 Trailing zeroes after the decimal point are significant.

(B. 3) 0.0473 Leading zeroes are not significant.

8 0
3 years ago
With 21 g of Zinc, and 7 g of CuCl2, how much ZnCl2 is made in grams?
mina [271]

Answer: 7.07 grams

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

\text{Moles of} zinc=\frac{21g}{65g/mol}=0.32moles

\text{Moles of} CuCl_2=\frac{7g}{134g/mol}=0.052moles

Zn+CuCl_2\rightarrow Cu+ZnCl_2

According to stoichiometry :

1 mole of CuCl_2 require 1 mole of Zn

Thus 0.052 moles of CuCl_2 will require=\frac{1}{1}\times 0.052=0.052moles  of Zn

Thus CuCl_2 is the limiting reagent as it limits the formation of product and Zn is the excess reagent.

As 1 mole of CuCl_2 give = 1 mole of ZnCl_2

Thus 0.052 moles of CuCl_2 give =\frac{1}{1}\times 0.052=0.052moles  of ZnCl_2

Mass of ZnCl_2=moles\times {\text {Molar mass}}=0.052moles\times 136g/mol=7.07g

Thus 7.07 g of ZnCl_2 will be produced from the given masses of both reactants.

8 0
3 years ago
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