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icang [17]
2 years ago
6

Is the double displacement reaction shown below, which of the products are insoluble

Chemistry
1 answer:
kati45 [8]2 years ago
8 0

Answer: AgCH

Explanation: I hope this helps

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A mixture of 0.682 mol of H2 and 0.440 mol of Br2 is combined in a reaction vessel with a volume of 2.00 L. At equilibrium at 70
ahrayia [7]

Answer: 0.294 mol of Br_2 present in the reaction vessel.

Explanation:

Initial moles of  H_2 = 0.682 mole

Initial moles of  Br_2 = 0.440 mole

Volume of container = 2.00 L

Initial concentration of H_2=\frac{moles}{volume}=\frac{0.682moles}{2.00L}=0.341M

Initial concentration of Br_2=\frac{moles}{volume}=\frac{0.440moles}{2.00L}=0.220M

equilibrium concentration of H_2=\frac{moles}{volume}=\frac{0.536mole}{2.00L}=0.268M

The given balanced equilibrium reaction is,

                            H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)

Initial conc.              0.341 M     0.220 M         0  M

At eqm. conc.    (0.341-x) M      (0.220-x) M     (2x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[HBr]^2}{[Br_2]\times [H_2]}

we are given : (0.341-x) = 0.268 M

x= 0.073 M

Thus equilibrium concentration of Br_2 = (0.220-x) M = (0.220-0.073) M = 0.147 M

[Br_2]=\frac{moles}{volume}\\0.147=\frac{xmole}{2.00L}\\\\x=0.294 mole

Thus there are 0.294  mol of Br_2 present in the reaction vessel.

7 0
3 years ago
Cuántos electrones contiene el átomo de helio​
Deffense [45]

Answer: 2:Dos

Explanation:

4 0
2 years ago
What produces hydrogen ions as the only positive ions in aqueous solution?
tatiyna
The answer is HB!

Hope this helps!
8 0
3 years ago
Read 2 more answers
Describe one way you could protect a bicycle from corrosion
Varvara68 [4.7K]
Don't use the brakes too hard when u stop. That can cause the biker to fall forward.

Give me a thanks if it helps!!
6 0
3 years ago
How many sulfur atoms are there in 21.0 g of al2s3?
ValentinkaMS [17]

Given the mass of aluminum sulfideAl_{2}S_{3} = 21.0 g

Molar mass of Al_{2}S_{3} = 2 * Molar mass of Al + 3 *molar mass of S = 2(27g/mol)+3(32.g/mol)=150.g/mol

Calculating the moles of Al_{2}S_{3}:

21.0 g *\frac{1 mol}{150 g} = 0.14 mol Al_{2}S_{3}

Each mole Al_{2}S_{3} has 3 mol S.

Each mole S constitutes 6.022*10^{23} atoms of S

Calculating atoms of S in 0.14 mol Al_{2}S_{3}:

0.14 mol Al_{2}S_{3} *\frac{3mol S}{1 mol Al_{2}S_{3}  }*\frac{6.022*10^{23}atoms S }{1 mol S}

= 2.53 * 10^{23} atoms of S

3 0
3 years ago
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