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Kobotan [32]
3 years ago
10

Balance this equation. If a coefficient of "1" is required, choose "blank" for that box. C2H6 + O2 → CO2 + H2O

Chemistry
2 answers:
Black_prince [1.1K]3 years ago
6 0

Explanation:

In a balanced equation, number of atoms on both reactant and product side are equal.

For example, C_{2}H_{6} + O_{2} \rightarrow CO_{2} + H_{2}O

Number of atoms present on reactant side are as follows.

C = 2

H = 6

O = 2

Number of atoms present on product side are as follows.

C = 1

H = 2

O = 3

Therefore, to balance this equation multiply C_{2}H_{6} by 2 and multiply O_{2} by 7 on reactant side. Whereas multiply CO_{2} by 4 and multiply H_{2}O by 6 on product side.

Therefore, the balanced chemical equation is as follows.

          2C_{2}H_{6} + 7O_{2} \rightarrow 4CO_{2} + 6H_{2}O

fenix001 [56]3 years ago
5 0
Step 1: Write the unbalanced equation,

                                 C₂H₆  +  O₂    →    CO₂  +  H₂<span>O

There are 2 C at left hand side and 1 carbon at right hand side. So, multiply CO</span>₂ by 2 to balance C atoms at both side. So,

                                 C₂H₆  +  O₂    →   2 CO₂  +  H₂O

Now, count number of H atoms at both sides. There are 6 H atoms at left hand side and 2 at right hand side. Multiply H₂O by 3 to balance H atoms.


                                 C₂H₆  +  O₂    →   2 CO₂  +  3 H₂O

At last, balance O atoms. There are 2 O atoms at left hand side and 3 O atoms at right hand side. Multiply O₂ with 1.5 (i.e. 3/2) to balance O atoms. i.e.

                                 C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O

Hence, the equation is balanced. If you want to make equation fraction free then multiply all equation with 2. i.e.

                           ( C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O ) × 2

                                2 C₂H₆  +  3 O₂    →   4 CO₂  +  6 H₂O
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The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0°c is 4.48. what is the value of ka for hbro?
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The solution would be like this for this specific problem:

Given:

 

pH of a 0.55 M hypobromous acid (HBrO) at 25.0 °C =  4.48

 

[H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-] <span>

Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9</span>

 

To add, Hypobromous Acid does not require acid adjustment, which is necessary for chlorine-based product and is stable and effective in pH ranges of 5-9.<span>

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Aluminum metal reacts with bromine, a red-brown liquid with a noxious odor. The reaction is vigorous and produces aluminum bromi
GuDViN [60]

<u>Answer:</u> The mass of bromine reacted is 160.6 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}        .....(1)

Given mass of aluminium = 18.1 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol

The chemical equation for the reaction of aluminium and bromide follows:

2Al+3Br_2\rightarrow 2AlBr_3

By Stoichiometry of the reaction:

2 moles of aluminium reacts with 3 moles of bromine gas

So, 0.670 moles of aluminium will react with = \frac{3}{2}\times 0.670=1.005mol of bromine gas.

Now, calculating the mass of bromine gas, we use equation 1:

Moles of bromine gas = 1.005 moles

Molar mass of bromine gas = 159.81 g/mol

Putting values in equation 1, we get:

1.005mol=\frac{\text{Mass of bromine}}{159.81g/mol}\\\\\text{Mass of bromine}=(1.005mol\times 159.81g/mol)=160.6g

Hence, the mass of bromine reacted is 160.6 grams.

5 0
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