Question

Chem help

Answer 1

@JustSaiyan

Answer 2

Alright.

Answer 3

Just hang on - Just finding it LOL

Answer 4

1. Determine the percent composition of the following compounds:
a. CuBr2
b. N2S2

Answer 5

A:
Cu: 28.45%
Br: 71.55%

Answer 6

Please show the full calculations on how you got that answer :)

Answer 7

Alright.

Answer 8

first determine the mass of each element

Answer 9

Count the number of hydrogens

Answer 10

Then, divide the elements to get the percent.

Answer 11

Alrighty, can you apply that into numbers now? :P

m = mass
h = hydrogen's
m/h?

Answer 12

1(63.546) + 2(79.904) = 223.354
63.546/223.354) X 100 = 28
Cu = 28%

{2(79.904) / 223.354} X 100 = 71.5
Br271.5%

That is A.

Answer 13

Br2 = 71.5%
or
Br271.5%? ._.

Answer 14

Br2 = 71.5%

Answer 15

*stare*

Answer 16

You've been typing a reply for a awefully long time

Answer 17

Determine the percent composition from the following analytical data:
a. An experiment was performed in which 6.1 g table salt produced 2.6 g Cl in the sample.
b. A reaction uses 18.06 grams Mg to react completely with 6.96 grams N

Answer 18

\[2 N atoms × \frac{ 14.01u }{ 1~n~atom } = 28.02 u.\]

\[ 2 S atoms × \frac{ 32.06u }{ 1~s~atom } = 64.12\]

\[\frac{ mass~of~N }{ total~mass } \times 100 % = \frac{ 28.02u }{ 92.14u } × 100 % =30.41 %\]

\[\frac{ mass~of~S }{ total~mass } \times 100 % = \frac{ 64.12u }{ 92.14u } × 100 % =69.59 %\]

Answer 19

Are you kidding. hold on. the last 2 ***ed up at the percent mark.

Answer 20

Ok o-o

Answer 21

\[\frac{ 28.02 u }{ 92.14 u } × 100~percent =30.41~percent\] End of ***ed up line one.

\[\frac{ 64.12 u }{ 92.14 u } × 100~percent =69.59~percent\] End of ***ed up line 2

Answer 22

Remember these are finishing the bottom to on the previous one that I said I ***ed up.

Answer 23

Damn that was a pain in the ***.

Answer 24

Is that all?

Answer 25

nope

Answer 26

What is a hydrate?

Answer 27

a substance that contains water or part of its elements.

Answer 28

Calculate the percent by mass of water in the following compounds:
a. FeCl3⋅6H2O
b. Ba(OH)2⋅8H2O

Answer 29

A.
Fe: 1 x 55.8 = 55.8
Cl: 3 x 35.45 = 106.4
H: 12 x 1.0 = 12.0
O: 6 x 16 = 96.0
The total is 270.2 g/mol.

% H2O = (108/270.2) x 100 = 40.0%

Answer 30

B.
Ba(OH)2 * 8 H2O = 315.46 g /mol
molar mass H2O = 18.02 g/mol
% water = 8 x 18.02 x 100/ 315.46 = 45.7

Answer 31

Calculate the empirical formula of each compound from the data listed below:
a) 65.5% C, 5.5% H, 29.0% O
b) 34.95% C, 6.844% H, 46.56% O, 13.59% N

Answer 32

the 65.5 g is carbon
5.5 g in hydrogen
and 29 grams is oxygen

C = 12.011 g/mole
H = 1.0079 g/mole
O = 15.9994 g/mole

moles C = 65.5g * mole/12.011g = 5.4533 moles
moles H = 5.5g * mole/1.0079 g = 5.4569 moles
moles O = 29g * mole/15.9994g = 1.8126 moles

divide by the lowest (Oxygen)

C = 5.4533/1.8126 = 3
H = 5.4569/1.8126 = 3
C = 1

Empirical formula = C3H3O MW = 55.0561 g/mole

Answer 33

Find the number of moles of each substance
C = 34.95 / 12 = 2.9125
H = 6.844/1 = 6.844
O = 46.56/16 = 2.91
N = 13.59/14 = 0.9707

Now you take the smallest number and divide into all the others, but including itself.

C = 2.9125/0.9707 = 3
H = 6.844 / 0.9707 = 7
O = 3 (same as the carbon
N = 1

The empirical formula = C3 H7 O3 N

Answer 34

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