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VashaNatasha [74]
2 years ago
13

Pls help i am in 8th grade k12

Chemistry
1 answer:
padilas [110]2 years ago
3 0
Imma go with explain, tell me if it’s wrong or right!!
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What is the empirical formula for a substance that is composed of 40.66% carbon, 8.53% Hydrogen,23.72% Nitrogen, and 27.09% Oxyg
prohojiy [21]

Answer:

THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO

Explanation:

The steps involved in calculating the empirical formula of this substance in shown in the table below:

Element                            Carbon           Hydrogen          Nitrogen         Oxygen

1. % Composition              40.66              8.53                 23.72                27.09

2. Mole ratio =

%mass/ atomic mass       40.66/12         8.53/1          23.72/14            27.09/16

                                       =  3.3883            8.53              1,6943             1.6931

3. Divide by smallest

value (0.6931)          3.3883/1.6931    8.53/1.6931    1.6943/1.6931   1.6931/1.6931

                               =      2.001                  5.038           1.0007                      1

4. Whole number ratio        2                       5                   1                               1

The empirical formula = C2H5NO

7 0
3 years ago
ANSWER ASAP PLEASE The Florida panther is one of the most endangered animals on earth. Where is the air that Florida panthers br
lutik1710 [3]
A. in the atmosphere
7 0
3 years ago
The formula for a different carbonate compound is KCO3.
vovangra [49]

Answer:

ω∉1

Explanation:

4 0
3 years ago
Why is the number of electrons or charge of each ion useful information when predicting the products in replacement reactions?
olga nikolaevna [1]
In order to determine if the ion is positively charged or negatively charged
3 0
3 years ago
What is the pressure of 0.33 moles of nitrogen gas, if its volume is 15.0 L at –25.0oC?
musickatia [10]

Using the ideal gas law PV =nRTPV=nRT , we find that the pressure will be P =\frac{nRT}{V}P=

V

nRT

​

 . Then, we'll substitute and find the pressure, using T = -25 °C = 248.15 K and R = 0.0821 \frac{atm\cdot L}{mol \cdot K}

mol⋅K

atm⋅L

​

 :

P =\frac{nRT}{V} = \frac{(0.33\,\cancel{mol})(0.0821\frac{atm\cdot \cancel{L}}{\cancel{mol \cdot K}})(248.15\,\cancel{K})}{15.0\,\cancel{L}} = 0.4482\,atmP=

V

nRT

​

=

15.0

L

​

(0.33

mol

)(0.0821

mol⋅K

atm⋅

L

​

​

)(248.15

K

​

)

​

=0.4482atm

In conclusion, the pressure of this gas is P=0.4482 atm.

Reference:

Chang, R. (2010). Chemistry. McGraw-Hill, New York.

3 0
2 years ago
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