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Jlenok [28]
2 years ago
11

If you have 23 g of Lithium oxide, what is the volume of Oxygen that is produced? _____ Li2O2 + _____ H2O  _____ LiOH + _____ O

2
Chemistry
1 answer:
Svet_ta [14]2 years ago
8 0

Answer:

1st step : find the number of moles for the known substance

number of moles =mass/ relative molecular mass

n=23/Li2O2

n=23/(14+32)

n =23/46

n= 0,5moles

2nd step : balance the equation

2Li202 +2H20-----> 4LiOH +O2

3rd step : find the mole ratio of the known or stated using the balanced equation

Li2O2 : O2

2 : 1

4th step :find the number of moles for the asked using the mole ratio using simple proportion

Li2O2 :O2

2 :1

0,5 :???

0,5/2 ×1

= 0,25moles for (O2)

5th step : find the asked either mass , volume , concentration

number of moles = volume/molar gas volume

volume = number of moles × molar gas volume

v = 0,25moles × 28dm^3

v= 7 dm^3 of (O2)

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Which electronic configuration matches the model?
antiseptic1488 [7]

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Read 2 more answers
How many molecules are in 145.5 grams of Be(OH)2
Zanzabum

Answer:

2.04 x 10²⁴ molecules

Explanation:

Given parameters:

Mass of Be(OH)₂ = 145.5g

To calculate the number of molecules in this mass of Be(OH)₂ we follow the following steps:

>> Calculate the number of moles first using the formula below:

Number of moles = mass/molarmass

Since we have been given the mass, let us derive the molar mass of Be(OH)₂

Atomic mass of Be = 9g

O = 16g

H = 1g

Molar Mass = 9 + 2(16 + 1)

= 9 + 34

= 43g/mol

Number of moles = 145.5/43 = 3.38mol

>>> We know that a mole is the amount of substance that contains Avogadro’s number of particles. The particles can be atoms, molecules, particles etc. Therefore we use the expression below to determine the number of molecules in 3.38mol of Be(OH)₂:

Number of

molecules= number of moles x 6.02 x 10²³

Number of molecules= 3.38 x 6.02 x 10²³

= 20.37 x 10²³ molecules

= 2.04 x 10²⁴ molecules

3 0
2 years ago
What is the molarity of a naoh solution if 39.1 ml of a 0.112 m h2so4 solution is required to neutralize a 25.0-ml sample of the
aleksandrvk [35]
The reaction between NaOH and H₂SO₄ is as follows;
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry  of base to acid is 2:1
NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place.
number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol
the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol
Number of NaOH moles in 25.0 mL - 0.0088 mol
Therefore in 1000 mL  - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L
Therefore molarity of NaOH - 0.352 M

8 0
3 years ago
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