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adell [148]
2 years ago
7

Consider the reaction; Mg + O2 → MgO Calculate the mass of magnesium oxide possible if 2.5 g Mg reacts with 5.0 g O2?

Chemistry
1 answer:
MAXImum [283]2 years ago
7 0

Answer:

4.15g

Explanation:

Balance the reaction:

2Mg + O2 = 2MgO

Find limiting reactant from the two, by dividing miles with coefficient

2.5/24.3 = 0.103 moles/2 = 0.0515

5/16 = 0.31 moles/2 = 0.156

0.0515 is small than 0.156, so limiting reactant is Mg

Compare the ratio of Mg with MgO

Ratio is 2:2

So same moles = 0.103 moles

Mass = moles x molar mass of MgO

Mass = 0.103 x 40.3

= 4.15g

*please give brainlest*

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Find a mole of 0.0960 g of H2SO4
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Molar mass H₂SO₄ = 98.079 g/mol

1 mol -------- 98.079 g
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= 0.0960 / 98.079

= 9.788 x 10⁻⁴ moles

hope this helps!
3 0
4 years ago
Based on the graph below, if you chose an alien at random on day 20, what form would it most likely be in?
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Answer:

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Explanation:

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4 0
3 years ago
For each of the following pairs of complexes, identify which one you would predict to have the larger Δo value, and explain why.
mash [69]

Answer:

a) [Fe(H2O)6]3+

b) [Fe(CN)6]3−

c) [Ru(CN)6]3-

Explanation:

. [Mn(H2O)6]2+ or [Fe(H2O)6]3+

The both complexes are d5 complexes with the same ligand , water. Water is a weak ligand and note that Mn^2+ often have a crystal field stabilization energy of zero hence

[Fe(H2O)6]3+ will possess a greater ∆o value.

The splitting of d orbitals according to the crystal field theory depends on the;

i)geometry of the complex

ii) nature of the metal ion,

iii)charge on the metal ion,

iv) ligands that surround the metal ion.

When the geometry and the ligands are held constant, the order of crystal field splitting is as follows;

Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+

[Fe(H2O)6]3+ or [Fe(CN)6]3−

[Fe(CN)6]3− will have a greater ∆o because the cyanide ion is a strong field ligand compared to water. A strong field ligand causes a greater splitting of the octahedral crystal field compared to a weak field ligand.

. [Fe(CN)6]3− or [Ru(CN)6]3-

[Ru(CN)6]3- will exhibit a greater crystal field splitting. Crystal field splitting increases with the second and third row transition elements when compared to the crystal field splitting of the first row transition elements. Note that, there is an increase of approximately 30%–50% in Δo on going from a first-row transition metal to a second-row metal and another 30%–50% increase on going from a second-row to a third-row metal when they have the same geometry and oxidation state.

4 0
3 years ago
Viết các đồng phân cấu tạo mạch hở của C4H6O2 cùng nhóm chức axit
Aleksandr-060686 [28]

Answer:

+ axit

CH2=CH-CH2-COOH,

CH3-CH=CH-COOH (tính cả đồng phân hình học)

CH2=C(CH3)-COOH.

+ este

HCOOCH=CH-CH3 (tính cả đồng phân hình học)

HCOO-CH2-CH=CH2,

HCOOC(CH3)=CH2.

CH3COOCH=CH2

CH2=CH-COOCH3

8 0
3 years ago
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