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Ne4ueva [31]
3 years ago
11

What is percentage composition?

Chemistry
1 answer:
blondinia [14]3 years ago
7 0

Answer:

The percent composition of a component in a compound is the percent of the total mass of the compound that is due to that component. To calculate the percent composition of a component in a compound: Find the molar mass of the compound by adding up the masses of each atom.

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Nearly all the energy from the earth receives from the sun is used in photosynthesis
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Answer: It is false because we use alot of it to with solar power.

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An electron in a hydrogen atom moves from level 3 to level 1. In a second hydrogen atom, an electron drops from level 2 to level
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I think the answer that you are looking for is B but im not a 100% sure

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In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions)
jeyben [28]

Answer:

Li2S> Na2S> K2S> CsS

Explanation:

The lattice energy of ionic species depends on the relative sizes of ions in the ionic compounds. As the size of ions increases, the lattice energy decreases and vice versa.

When the size of the anions are the same, the lattice energy now depends on the relative sizes of the cations. Therefore, since all the compounds are sulphides and the order of magnitude of ionic sizes is: Li^+ < Na^+ < K^+ < Cs^+.

Therefore, the order of decrease in lattice energy is; Li2S> Na2S> K2S> CsS

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3 years ago
The process of_ causes rock to change compostion when reacting with oxgen
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Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given
professor190 [17]

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]

-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]

\Delta H_{C_2H_2}=-227kJ

Therefore, the enthalpy change for C_2H_2 is -227 kJ.

7 0
3 years ago
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