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Korolek [52]
3 years ago
12

How many moles are in 17.50g of SO2

Chemistry
1 answer:
monitta3 years ago
8 0
The formula for mole is
n= Mass/Mol mass
Mol Mass: S=32
O2= 16(2)
—————
64 g/mol
N= 17.50 g
————— (cancel both g)
64 g/mol
= 0.27 mol is the answer
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Given a Ksp for AgBr of 5.0 * 10–13, what happens when 50 ml of 0.002 M AgNO3 and 50 mL of 0.002 M NaBr are mixed?
IrinaK [193]

Answer:

A precipitate will be produced

Explanation:

The Ksp of AgBr is:

AgBr(s) → Ag⁺ + Br⁻

5.0x10⁻¹³ = [Ag⁺] Br⁻]

<em>Where [] are the concentrations in equilibrium of each ion.</em>

<em />

And if Q is:

Q = [Ag⁺] Br⁻]

<em>Where the concentrations are actual concentrations of each ion</em>

<em />

We can say:

IF Q >= Ksp, a precipitate will be produced

IF Q < Ksp, no precipitate will be produced.

the molar concentrations are:

[AgNO₃] = [Ag⁺] = 0.002M * (50mL / 100mL) = 0.001M

<em>Because 50mL is the volume of the AgNO₃ solution and 100mL the volume of the mixture of both solutions.</em>

[NaBr] = [Br⁻] = 0.002M * (50mL / 100mL) = 0.001M

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5 0
3 years ago
Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9.30x10⁻⁸ at 700°C:2H₂S(g) ⇄ 2H₂(g) + S₂(g) If 0
astra-53 [7]

the equilibrium concentration of H₂(g) at 700°C =  0.00193 mol/L

0.00193 mol/L

Given that:

numbers of moles of H₂S = 0.59 moles

Volume = 3.0-L

Equilibrium constant  = 9.30 × 10⁻⁸

The equation for the reaction is given as :

2H₂S    ⇄   2H₂(g)  +  S₂(g)

The initial concentration of H₂S =

The initial concentration of H₂S =

= 0.1966 mol/L

The ICE table is shown be as :

                           2H₂S                ⇄         2H₂(g)        +        S₂(g)

Initial                    0.9166                           0                           0

Change                 -2 x                               +2 x                      + x

Equilibrium          (0.9166 - 2x)                   2x                         x

     

(since 2x < 0.1966 if solved through quadratic equation)

The equilibrium concentration for H₂(g) = 2x

∴

= 0.00193 mol/L

Thus, the equilibrium concentration of H₂(g) at 700°C =  0.00193 mol/L

To know more about equilibrium concentration

brainly.com/question/13414142

#SPJ4

5 0
1 year ago
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