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Salsk061 [2.6K]
2 years ago
5

If a sample of magnesium with a mass of 35.0 grams reacts with 35.0 grams of oxygen how much magnesium oxide will be produced?​

Chemistry
1 answer:
slamgirl [31]2 years ago
8 0

Answer:

58.0 g of MgO

Explanation:

in a perfect world, 70 g, however we don't live in a perfect world

The equation of reaction

2Mg + O₂ --> 2MgO

first find which element is limiting:

35 g x 1 mol/24.3 g of Mg x 2 mol of MgO/ 2 mole of Mg = 1.44 moles of MgO

35 g x 1 mol/32g of Mg x 2 mol of MgO/ 1 mole of O₂ = 2.1875 moles of MgO

This means Mg is the limiting factor, so you will be using this moles to find grams of MgO

1.44 mols of MgO x 40.3 g of MgO/ 1 mol = 58.0 g of MgO

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Uranium (VIII) Sulfide formula
tresset_1 [31]

Answer:

US₂

Explanation:

Uranium sulfide (US₂)

Uranium atomic symbol = U

Sulfur atomic symbol = S

Uranium valency = +4

Sulfur valency = -2

So;

Uranium sulfide (US₂)

8 0
3 years ago
I’d appreciate the help! :)
seraphim [82]

Answer: 300g

Explanation:

first we write the given values on top

224L. x

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then we form a formula

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3 years ago
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5 0
3 years ago
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Explain the difference between weathering and erosion.
yanalaym [24]

Answer:

So, if a rock is changed or broken but stays where it is, it is called weathering. If the pieces of weathered rock are moved away, it is called erosion.

6 0
2 years ago
Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the follo
siniylev [52]

Answer:

∆H > 0

∆Srxn <0

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Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.

3 0
3 years ago
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