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pickupchik [31]
3 years ago
11

2) Find the volume that 2.0 moles of H2 will occupy at STP?

Chemistry
1 answer:
Rom4ik [11]3 years ago
4 0

Answer:

V = 44.85 L

Explanation:

Given data:

Volume of H₂ = ?

Number of moles of H₂ = 2.0 mol

Given temperature = 273.15 K

Given pressure = 1 atm

Solution:

Formula:

PV = nRT

P = Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

By putting values,

1 atm × V = 2.0 mol × 0.0821 atm.L/ mol.K  × 273.15 K

V = 44.85 atm.L / 1 atm

V = 44.85 L

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The balanced chemical reaction for this would be written as:

2Mg + O2 = 2MgO

We use this reaction and the amount of the reactant given to calculate for the amount of magnesium oxide that is produced. We do as follows:

1.5 g Mg (1 mol / 24.31 g) ( 2 mol MgO / 2 mol Mg ) (40.30 g /1 mol ) = 2.49 g MgO produced
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Which best describes an element?
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6 0
3 years ago
g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron
Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ →  Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0
3 years ago
The cubic centimeter (cm3) is a measurement of which of the following quantities? Density
Murljashka [212]

(cm3)  is a commonly used unit of volume that extends the derived SI-unit cubic metre, and corresponds to the volume of a cube that measures 1 cm × 1 cm × 1 cm.

so the answer is volume



8 0
2 years ago
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