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Nana76 [90]
2 years ago
11

The first step in the process used to recover zinc metal from zinc sulfide ore is the reaction of zinc sulfide with oxygen gas t

o produce zinc oxide and sulfur dioxide. 2ZnS(s) 3O2(g)⟶2ZnO(s) 2SO2(g) When the external pressure is 1.523×105 Pa and the temperature is 675 K, the amount of work performed is 850.1 J. Calculate how many grams of oxygen are consumed in the reaction.
Chemistry
1 answer:
Vikentia [17]2 years ago
4 0

Answer : The mass of oxygen consumed in the reaction is 14.6 grams.

Explanation :

First we have to calculate the change in moles of oxygen.

Formula used :

w=\Delta nRT

or,

\Delta n=\frac{w}{RT}

where,

w = work done = 850.1 J

\Delta n = change in moles of gas = ?

R = gas constant = 8.314 J/mol.K

T = temperature = 675 K

Now put all the given values in the above formula, we get:

\Delta n=\frac{850.1J}{(8.314J/mol.K)\times (675K)}=0.152mol

Now we have to calculate the mass of oxygen consumed.

The balanced chemical reaction is,

2ZnS(s)+3O_2(g)\rightarrow 2ZnO(s)+2SO_2(g)

Here, the volume changes of solids can be ignored.

From the balanced chemical reaction, we conclude that

The number of moles oxygen required for the complete reaction = 3

Change in mole of oxygen = 0.152 mol

Molar mass of O_2 = 32 g/mol

So,

The mass of oxygen consumed = 3 x 0.152 mol x 32 g/mol = 14.6 g

Therefore, the mass of oxygen consumed in the reaction is 14.6 grams.

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If 45.5 g of a metal that has a density of 3.65 g/mL is placed in 45.0 mL of water, what is the final volume?
mart [117]

Answer:

The final volume of the metal and water is 54.46mL

Explanation:

Hello,

To solve this question, we'll first of all find the volume of the metal and assuming there's no loss of water by overflow in the container, we'll add the volume of the metal to the volume of the water to get the final volume.

Data;

Mass of the metal = 45.5g

Volume of the water = 45mL

Density of the metal (ρ) = 3.65g/mL

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ρ = mass / volume

Volume (v) = mass / density

Volume = 45.5 / 3.65

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The volume of the metal is 12.46mL.

When the metal is added to the container 45mL of water, assuming no water was lost by overflow in the container, the final volume =

Final volume = volume of metal + volume of water

Final volume = 12.46 + 45.0

Final volume = 57.46mL

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2 years ago
If the pOH of vinegar is 9.45, what is its [OH− ]?
7nadin3 [17]
Answer:
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Solution:

pOH is related to [OH⁻] as,

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Putting value of pOH,

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Solving for [OH⁻],

                           [OH⁻]  =  10⁻⁹·⁴⁵                       ∴ 10  =  Antilog

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