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otez555 [7]
3 years ago
15

What does the law of convservation of matter show?​

Chemistry
1 answer:
velikii [3]3 years ago
3 0
The law of conservation of energy states that energy is never created or destroyed only converted.
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How many valence electrons are available of bonding in bromine(Br)?
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 there are seven valence electrons<span> available of bonding in bromine(Br)</span>
5 0
3 years ago
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What is the theoretical yield of vanadium, in moles, that can be produced by the reaction of 1.0 mole of V2O5 with 4.0 moles of
IRINA_888 [86]

Answer:

Theoretical yield of vanadium = 1.6 moles

Explanation:

Moles of V_2O_5 = 1.0 moles

Moles of Ca = 4.0 moles

According to the given reaction:-

V_2O_5_{(s)} + 5Ca_{(l)}\rightarrow 2V_{(l)} + 5CaO_{(s)}

1 mole of V_2O_5 react with 5 moles of Ca

Moles of Ca available = 4.0 moles

Limiting reagent is the one which is present in small amount. Thus, Ca is limiting reagent. (4.0 < 5)

The formation of the product is governed by the limiting reagent. So,

5 moles of Ca on reaction forms 2 moles of V

1 mole of Ca on reaction for 2/5 mole of V

4.0 mole of Ca on reaction for \frac{2}{5}\times 4 mole of V

Moles of V = 1.6 moles

<u>Theoretical yield of vanadium = 1.6 moles</u>

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3 years ago
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I don’t really have a clear answer for you but this might help

7 0
3 years ago
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Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

5 0
3 years ago
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