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ICE Princess25 [194]
3 years ago
7

Balance and find out the coefficients of reactions and products ​

Chemistry
1 answer:
kolezko [41]3 years ago
6 0

Answer:

2, 1, 1, 4.

Explanation:

Hello there!

In this case, for the given chemical reaction:

H_2SO_4+Pb(OH)_4\rightarrow Pb(SO_4)_2+H_2O\\

We can see how there is one SO4 on the left and two on the right, thus, we add a 2 in front of H2SO4:

2H_2SO_4+Pb(OH)_4\rightarrow Pb(SO_4)_2+H_2O\\

Next, since there are 8 atoms of hydrogen on the left and two on the right, we add a 4 in front of H2O to obtain:

2H_2SO_4+Pb(OH)_4\rightarrow Pb(SO_4)_2+4H_2O\\

Which is now balanced so the coefficients 2, 1, 1, 4.

Best regards!

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A compound has a percent composition of 40.0% carbon , 6.72% hydrogen , and 53.28% oxygen . If it's molar mass is 180 g/mol , wh
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1) Molar composition

Base: 100 g

C: 40.00 g / 12.0 g /mol = 3.333 mol

H: 6.72 g / 1 g/mol = 6.72 mol

O: 53.28 g / 16g/mol = 3.33 mol

2) Divide by the smaller number of moles

C: 3.333 / 3.33 = 1.00

H: 6.72 / 3.33 = 2.02 ≈ 2

O: 3.33/3.33 = 1.00

3) Empirical formula

CH2O

molar mass of the empirical formula = 12 g/mol + 2*1g/mol + 16 g/mol = 30 g/,ol

4) Number of times that the mass of empirical formula is contained in the molar mass = 180 g/mol / 30 g/mol = 6

5) Molecular formula = 6 times the empirical formula

=>C6H12O6

Answer: C6H12O6

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Explanation:

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Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p
UNO [17]

Answer : The equilibrium constant K_c for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of Br_2.

\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}

\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M

Now we have to calculate the dissociated concentration of Br_2.

The balanced equilibrium reaction is,

                              Br_2(g)\rightleftharpoons 2Br(aq)

Initial conc.         1.731 M      0

At eqm. conc.      (1.731-x)    (2x) M

As we are given,

The percent of dissociation of Br_2 = \alpha = 1.2 %

So, the dissociate concentration of Br_2 = C\alpha=1.731M\times \frac{1.2}{100}=0.2077M

The value of x = 0.2077 M

Now we have to calculate the concentration of Br_2\text{ and }Br at equilibrium.

Concentration of Br_2 = 1.731 - x  = 1.731 - 0.2077 = 1.5233 M

Concentration of Br = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

K_c=\frac{[Br]^2}{[Br_2]}

Now put all the values in this expression, we get :

K_c=\frac{(0.4154)^2}{1.5233}=0.1133

Therefore, the equilibrium constant K_c for the reaction is, 0.1133

7 0
3 years ago
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