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fiasKO [112]
3 years ago
14

In the reaction ___h2+n2 ->2 NH3, what coefficient should be placed in front of H2 to balance the reaction?

Chemistry
2 answers:
Nataly_w [17]3 years ago
7 0
In order to balance this equation you need to count each element and how many of the individual elements are in the equation.
_H2+N2=2 NH3
You multiply the 2 (Which is the coefficient) by the 3 (which is the subscript) This would equal 6 which indicated there are 6 hydrogen atoms on the right side so the left side should also have 6 hydrogen atoms

The missing coefficient on the left side must multiple the 2 to become 6 hydrogen
Answer=3
Umnica [9.8K]3 years ago
7 0
The answer is 3, I just took the Test
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My Teacher didn't teach us this please help
oksian1 [2.3K]

Answer:

first; why is it a question if teacher never teached you it LOL anyways..

Explanation:

i would say its the second one

"The smae quality of each element is present on both sides of the equation."

5 0
3 years ago
5. Rocks and sediment change their form as they move through the rock cycle. Which change
mina [271]

Answer:

Rock to Magma

Explanation:

Lava to Rock is incorrect because its "Freezing" The Lava

Magma to Lava basically a name change. No type change.

Rock to Sediment is Erosion of the rock.

Rock to Magma Is the Rock being super-heated, melting it. Similar to Ice and water but more extreme.

8 0
3 years ago
Please help! I will give brainilist........
Annette [7]

Answer:

IT IS D because lights are heated and it gives light!

4 0
2 years ago
A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added
Ulleksa [173]

<u>Answer:</u> The pH change of the buffer is 0.30

<u>Explanation:</u>

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})

pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})        .....(1)

We are given:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.306M

[C_6H_5NH_2]=0.418M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH_{initial}=14-8.99=5.01

To calculate the molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M

The chemical reaction for aniline and HCl follows the equation:

                   C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-

<u>Initial:</u>           0.418        0.124           0.306

<u>Final:</u>             0.294          -                0.430

Calculating the pOH by using using equation 1:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.430M

[C_6H_5NH_2]=0.294M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH_{final}=14-9.29=4.71

Calculating the pH change of the solution:

\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30

Hence, the pH change of the buffer is 0.30

8 0
3 years ago
Analysis of a compound of sulfur, oxygen and fluorine showed that it is 31.42% S and 31.35% O, with F accounting for the remaind
Leya [2.2K]

Answer:

The molecular formula is  SO2F2

Explanation:

Step 1: Data given

Suppose the mass of compound = 100 grams

The compound contains:

31.42 % S = 31.42 grams S

31.35 % O = 31.35 grams O

100 - 31.42 - 31.35 = 37.23 F

Molar mass of S = 32.065 g/mol

Molar mass F = 19.00 g/mol

Molar mass O = 16.00 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles S = 31.42 grams / 32.065 g/mol

Moles S = 0.9799 moles

Moles 0 = 31.35 grams / 16.00 g/mol

Moles 0 = 1.959 moles

Moles F = 37.23 grams / 19.00 g/mol

Moles F = 1.959 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.9799 / 0.9799 = 1

F: 1.959/ 0.9799 = 2

O : 1.959 / 0.9799 = 2

The empirical formula is SO2F2

This formula has a molecular mass of 102.06 g/mol

This means the empirical formula is also the molecular formula : SO2F2

3 0
3 years ago
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