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Rina8888 [55]
3 years ago
15

Help pleaseee

Chemistry
1 answer:
marissa [1.9K]3 years ago
6 0

Answer:

In order to find the molecular formula from an empirical formula you must find the ratio of their molecular masses.

We know that the molecular mass of the molecule is 70

gmol-1

. We can calculate the molar mass of

CH2

from the periodic table:

C=12.01

gmol−1

H=1.01

gmol−1

CH2 =14.03

gmol−1

Hence we can find the ratio:

14.03

70

≈

0.2

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When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coeffici
Sergio [31]

<u>Answer:</u> The molecular balanced equation is written below

<u>Explanation:</u>

A molecular equation is defined as the chemical equation in which the ionic compounds are written as molecules rather than component ions.

A balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is equal to the total number of individual atoms on product side.

The balanced chemical equation for the decomposition of bromine trifluoride follows:

2BrF_3\rightarrow Br_2+3F_2

By Stoichiometry of the reaction:

2 moles of bromine trifluoride produces 1 mole of bromine gas and 3 moles of fluorine gas

Hence, the molecular balanced equation is written above.

7 0
3 years ago
If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha
NARA [144]
There are 2.32 x 10^6 kg sulfuric acid in the rainfall. 

Solution: 
We can find the volume of the solution by the product of 1.00 in and 1800 miles2: 
     1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m 
     1.00 in * 1 m / 39.3701 in = 0.0254 m  
     Volume = 4.662 x 10^9 m^2 * 0.0254 m
                  = 1.184 x 10^8 m^3 * 1000 L / 1 m3
                  = 1.184 x 10^11 Liters 

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70: 
     [H+] = 10^-pH = 10^-3.7 = 0.000200 M 
     [H2SO4] = 0.000100 M  
 
By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid: 
     1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4 

We can now calculate for the mass of sulfuric acid in the rainfall: 
     mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
                               = 2.32 x 10^9 g * 1 kg / 1000 g
                               = 2.32 x 10^6 kg H2SO4
3 0
4 years ago
Read 2 more answers
The density of ethanol, C2H5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 15.0 g of CO2 via a combust
bearhunter [10]

<u>The answer is </u><u>9.94 ml.</u>

<h3>What is density?</h3>
  • Density is a word we use to describe how much space an object or substance takes up (its volume) in relation to the amount of matter in that object or substance (its mass).
  • Another way to put it is that density is the amount of mass per unit of volume. If an object is heavy and compact, it has a high density.

Given,

        The density of ethanol, C2H5OH = 0.789 g/mL

n (CO_{2} ) = \frac{m}{M}  = \frac{15G}{44 g/mol} } = 0.341 mol;

n ( C_{2} H_{2} OH) = \frac{n (CO_{2}) }{2}  = \frac{0.341}{2}  = 0.1705  mol;

m (C_{2} H_{2} OH) = 0.1705 mol * 46  g/ mol = 7.843 g

V (C_{2} H_{2} OH ) = \frac{7.843}{0.789} = 9.94 ml.

Therefore, the answer is 9.94 ml

Learn more about density of ethanol,

brainly.com/question/18597444

#SPJ4

<u>The complete question is -</u>

If the density of ethanol, C2H5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 15.0 g of CO2 according to the following chemical equation?

C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

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2 years ago
What is the general formula for a secondary amine? rnh2 h2nrnh2 rn2h r2nh
zvonat [6]

The general fomula of the amine group ought to be R2NH because, the two hydrogen atoms of ammonia have been substituted by alkyl groups.

<h3>What is a secondary amine?</h3>

A secondary amine is one in which two alkyl groups are subtituted on the nitrohen atom of ammonia.

This implies that the general fomula of the amine group ought to be R2NH. This is because, the two hydrogen atoms of ammonia have been substituted by alkyl groups.

Learn more about secondary amine: brainly.com/question/12682525

4 0
2 years ago
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katrin [286]
The correct answer is:  [C]:  " mg " {"milligrams"} .
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4 0
3 years ago
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