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SVETLANKA909090 [29]
3 years ago
10

2H2 +O2 = 2H2O balance equation with 5 oxygen

Chemistry
1 answer:
Ivenika [448]3 years ago
5 0

10H₂    +   5O₂ →    10H₂O

Explanation:

This problem deals with balancing of chemical equations. In balancing chemical equations, the law of conservation of mass must be followed. This states that:

   "In a chemical reaction, matter is neither created nor destroyed but transformed from one form to another".

This meaning of this is that; the number of atoms on each side of the expression must be the same.

                    2H₂    +   O₂ →    2H₂O

let us check is the equation above is balanced;

   

                                      2H₂    +   O₂ →    2H₂O

Elements             reactant                product

H                                  4                           4

O                                   2                           2

We can see vividly that the equation is balanced;

Now; if we have 5 oxygen gas, we multiply the equation through by 5:

 

                                 5 x          (  2H₂    +   O₂ →    2H₂O   )

     ⇒        10H₂    +   5O₂ →    10H₂O

Elements             reactant                product

H                                  20                       20

O                                  10                        10

learn more:

Balanced equation brainly.com/question/11102790

#learnwithBrainly

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Ne4ueva [31]

Answer:

1.09 M

Explanation:

Let's define the equation that will be used to calculate the final concentration of the resultant calcium nitrate solution. In order to calculate it, we need to find the total number of moles of calcium nitrate and divide by the total volume of the resultant solution:

c=\frac{n}{V}

This equation firstly helps us find the number of moles of calcium nitrate. Multiplying molarity by volume will yield the moles. Adding the moles from the first component to the second component will provide us with the total number of moles of calcium nitrate:

n_{Ca(NO_3)_2}=46.2 mL\cdot0.568 M+80.5 mL\cdot1.396 M=138.62 mmol

Now, the total volume of this solution can be found by adding the volume values of each component:

V_total=46.2 mL+80.5 mL=126.7 mL

Finally, dividing the moles found by the total volume will yield the final molarity:

c_{final}=\frac{138.62 mmol}{126.7 mL}= 1.09 M

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PCl3(g) + Cl2(g) ⇋ PCl5(g) Kc = 91.0 at 400 K. What is the [Cl2] at equilibrium if the initial concentrations were 0.24 M for PC
Dmitry_Shevchenko [17]

Answer:

[Cl₂] in equilibrium is 1.26 M

Explanation:

This is the equilibrium:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Kc = 91

So let's analyse, all the process:

                PCl₃(g)        +        Cl₂(g)     ⇋        PCl₅(g)

Initially     0.24 M                 1.50M                 0.12 M

React           x                           x                         x

Some amount of compound has reacted during the process.

In equilibrium we have

              0.24 - x                  1.50 - x                  0.12 + x

As initially we have moles of product, in equilibrium we have to sum them.

Let's make the expression for Kc

Kc = [PCl₅] / [Cl₂] . [PCl₃]

91 = (0.12 + x) / (0.24 - x) ( 1.50 - x)

91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)          

91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)

32.76 - 158.34x + 91x² = 0.12 +x

32.64 - 159.34x + 91x² = 0

This a quadratic function:

a = 91; b= -159.34; c = 32.64

(-b +- √(b² - 4ac)) / 2a

Solution 1 = 1.5

Solution 2 = 0.23 (This is our value)

So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M

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Dmitry_Shevchenko [17]

Answer:

1 : 2 ratio

Explanation:

Chlorine has a coefficient of 1, and sodium chloride has a coefficient of 2.

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