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WARRIOR [948]
2 years ago
6

2MG+O2 ...........2MGO

Chemistry
1 answer:
Galina-37 [17]2 years ago
5 0
Mg + 1/2 O2 → MgO

1 mol = 24 g of Mg

X mol = 12 g of Mg

x = 0.5 moles of Mg

Mg :MgO = 1:1 (coefficient from equations using mole ratio)

So

0.5 moles of MgO

1 mol MgO = (24+16) g = 40 g

0.5 moles of MgO = 0.5 × 40

= 20 g of MgO produced
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How does the number of neutrons compare to the number of protons in a stable atom?
bulgar [2K]
Number of neutrons = Atomic mass - Atomic number (number of protons)
Hope that helps!
6 0
3 years ago
What are the units of molar mass?<br> A. L/g<br> B. mol/g<br> C. g/L<br> D. g/mol<br> SUBST
rosijanka [135]

Answer:

D - g/mol

step by step method

7 0
3 years ago
Read 2 more answers
Consider the reaction below that has a Keq of 8.98 X 10-2. The initial concentration of A and B is 1.68 M. Calculate the equilib
DaniilM [7]

Answer:

[C] = 0.4248M

Explanation:

                   A     +     B      ⇄     2C

C(i)          1.68M      1.68M         0.00

ΔC              -x            -x             +2x

C(eq)      1.68-x       1.68-x          2x

Keq = [C]²/[A][B] = (2x)²/(1.68 - x)²= 8.98 x 10⁻²

Take SqrRt of both sides => √(2x)²/(1.68 - x)² = √8.98 x 10⁻²

=> 2x/1.68 - x = 0.2895

=> 2x = 0.2895(1.68 - x)

=> 2x = 0.4863 - 0.2895x

=> 2x + 0.2895x = 0.4863

=> 2.2895x = 0.4863

=> x = 0.4863/2.2895 = 0.2124

[C] = 2x = 2(0.2124)M = 0.4248M in 'C'

3 0
3 years ago
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Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25∘C diamond changes to graphite so slowly that the entha
rjkz [21]

Answer:

-1.9 KJ/mol

Explanation:

In order to solve the problem, we have to rearrange the equations in a way in which all molecules of O₂ and CO₂ are eliminated:

2C(diamond) + 2O₂(g) → 2CO₂(g)     ΔH₁= 2 x (-395.4 KJ) ------> we multiply by 2 both reactants and products

2 CO₂(g) → 2CO(g) + O₂(g)         ΔH₂= 566.0 KJ

CO₂(g) → C(graphite) + O₂(g)     ΔH₃= -1 x (-393.5 KJ) ------> we use reverse rxn

2CO(g) → C(graphite) + CO₂(g)   ΔH₄= -172.5 KJ

When we cancel the molecules that appear both in reactants and products, the total reaction is the following:

2C(diamond) → 2C(graphite)

ΔHt= ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ = 2 x (-395.4 KJ) + 566.0 KJ + (-1 x (-393.5 KJ)) - 172.5 KJ

ΔHt= 347.2 KJ

This is for 2 mol of C(diamond) which are converted in 2 mol of C(graphite). To obtain ΔH for the reaction of 1 mol C(diamond) to 1 mol (graphite) we have to divide into 2:

ΔH= -3.8 KJ/2mol= -1.9 KJ/mol

5 0
3 years ago
Which is a set of valid quantum numbers?
harina [27]
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2,1,-1,1/2 -- Valid
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4,3,3,-1/2 -- Valid
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0,2,1,1/2 -- NOT (no such thing as n = 0, l too high)
2,0,0,-1/2 -- Valid
1,0,0,1/2 -- Valid
3,2,1,0 -- NOT (cannot have ms = 0)
4,3,4,-1/2 -- NOT (ml too high)

6 0
3 years ago
Read 2 more answers
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