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3 years ago

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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?

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