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Lunna [17]
3 years ago
10

Which of the following compounds results from the formation of an ionic bond?

Chemistry
1 answer:
Reptile [31]3 years ago
6 0

Answer:

my answer is a I think but i an not sure

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Which description gives the steps for creating hydroelectric power
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B) river water stored in a reservoir flows down a channel in a dam, turning a turbine as it passes. the turbine is connected to a generator that produces electricity, and then the water continues downstream

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Explanation:

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A material through which heat travels easily is what
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3 years ago
Sodium azide decomposes rapidly to produce nitrogen gas.
brilliants [131]

Answer:

156 g

Explanation:

Let's consider the following reaction.

2 NaN₃(s) → 2 Na(s) + 3 N₂ (g)

We can find the moles of N₂  using the ideal gas equation.

P × V = n × R × T

1.50 atm × 60.0 L = n × (0.08206 atm.L/mol.K) × 305 K

n = 3.60 mol

The molar ratio of N₂ to NaN₃ is 3:2. The moles of NaN₃ are:

3.60 mol N₂ × (2 mol NaN₃ / 3 mol N₂) = 2.40 mol NaN₃

The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:

2.40 mol × 65.01 g/mol = 156 g

4 0
3 years ago
C6H12O6 + 6O2 ---> 6H2O + 6CO2
Ivahew [28]

Answer:

24e⁻ are transferred by the reaction of respiration.

Explanation:

C₆H₁₂O₆   +  6O₂   →   6 H₂O   +  6CO₂

This is the reaction for the respiration process.

In this redox, oxygen acts with 0 in the oxidation state on the reactant side, and -2 in the product side -  REDUCTION

Carbon acts with 0 in the glucose (cause it is neutral), on the reactant side and it has +4, on the product side - OXIDATION

6C →  6C⁴⁺  +  24e⁻

In reactant side we have a neutral carbon, so as in the product side we have a carbon with +4, it had to lose 4e⁻ to get oxidized, but we have 6 carbons, so finally carbon has lost 24 e⁻

6O⁻² +  6O₂  + 24e⁻  →  6O₂²⁻  +  6O⁻²

In reactant side, we have 6 oxygen from the glucose (oxidation state of -2) and the diatomic molecule, with no charge (ground state), so in the product side, we have the oxygen from the dioxide with -2 and the oxygen from the water, also with -2 at the oxidation state. Finally the global charge for the product side is -36, and in reactant side is -12, so it has to win 24 e⁻ (those that were released by the C) to be reduced.

3 0
3 years ago
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