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4 years ago

A sample of a given compound contains 13.18g of carbon and 3.32g of hydrogen. What is the percent composition of the compound?

Chemistry

1 answer:

GarryVolchara [31]4 years ago

8 0

The percent composition of the compound is 79.87 % of carbon and 20.12% of hydrogen

Explanation:

Given:

mass of carbon = 13.18 grams

mass of hydrogen = 3.32 grams

percent composition of compound =?

From the given data total mass of the compound is calculated

13.8 + 3.32

= 16.5 grams

The individual percentage of hydrogen and carbon will be calculated as:

percent composition = $\frac{individual mass of the element}{total mass of the compound}$ x 100

For carbon,

$\frac{13.18}{16.5}$ x 100

percent composition of carbon = 79.87 % of carbon.

for hydrogen:

$\frac{3.32}{16.5}$

= 20.12 % of hydrogen

In the compound given 79.87 % of carbon and 20.12% of hydrogen is present.

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Complete and balance the following acid-base equations:

Alchen [17]

Answer:

a) $HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O$

b) $H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)$

c) $Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O$

Explanation:

when $HClO_4$ is added to LiOH, lithium chlorate and water is formed.

$HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O$

Balancing of above reaction,

It can be seen that all the atoms in both the sides are balanced. so it is a balanced reaction.

(b) When aqueous $H_2SO_4$ is added to NaOH, $Na_2SO_4$ and $H_2O$ is formed. It is a neutrilization reaction.

$H_2SO_4(aq) + NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)$

Balancing of above reaction,

First balance all the atoms except O and H

S atom is already balanced in either side

No. of Na atom in left hand side = 1

No. of Na atom in right hand side = 2

So multiply NaOH by 2, now the reaction becomes:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)$

No. of O atoms in left hand side = 6

No. of O atoms in right hand side = 5

So multiply H2O by 2, now the reaction becomes:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)$

Now, it can been seen that all the atoms are balanced.

c) When $Ba(OH)2$ reacts with HF, baroum fluoride and water is formed.

$Ba(OH)2 + HF\rightarrow BaF_2 + H_2O$

Balancing of above reaction,

Barium atoms are already balanced on either side.

No. of F atoms on left hand side = 1

No. of F atoms on right hand side = 2

So, multiply HF by 2, now reaction becomes

$Ba(OH)2 + 2HF\rightarrow BaF_2 + H_2O$

In order to balance H and O on either side, multiply H2O by 2.

$Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O$

7 0

3 years ago

What mass of cu(s) is electroplated by running 26.5 a of current through a cu2+(aq) solution for 4.00 h?

kramer

T = 14400 s
26.5 x 14400=381600 C
381600/96500=3.95 Faradays
Cu2+ + 2e- = Cu
3.95 faradays ( 1 mol/ 2 Faradays) = 1.97
mass = 1.97 x 63.55 g/mol=125 g

moles Au = 33.1 / 196.967 g/mol=0.168
Au+ + 1e- = Au
0.168 ( 1 Faraday/ 1mol)= 0.168 Faraday
0.168 x 96500=16217 Coulombs
16217 / 5.00=3243 s => 54 min

7 0

3 years ago

. Write the following isotope in nuclide notation (e.g., “ ”): copper-70

vekshin1

Answer:

${\boxed{\text{$_{29}^{70}${Cu}}}$

Explanation:

The atomic number (Z) of copper is 29 and this isotope has an atomic mass (A) of 70.

The general symbol for an isotope E is $_{Z}^{A}\text{E}$ .

The atomic number is a left subscript, and the atomic mass is a left superscript.

$\rm {\text{The nuclide notation for copper-70 is }}{\boxed{\textbf{$_{29}^{70}${Cu}}}$

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3 years ago

One end of a molecule has a positive charge and the other end of the same molecule has a negative charge. What is the best term

klasskru [66]

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OverLord2011 [107]

Letter B is correct the answer

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