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antoniya [11.8K]
3 years ago
15

Explain why solid sodium chloride does not conduct electricity

Chemistry
1 answer:
aleksandr82 [10.1K]3 years ago
8 0

Answer:

It does not contain free mobile ions

Explanation:

Solid sodium chloride does not conduct electricity because it does not contain free mobile ions.

  • Although sodium chloride is an ionic compound, in solid state, there is absence of free mobile ions.
  • In aqueous solutions or in molten form, ionic compounds can conduct an electric current using free mobile ions as carriers.
  • Therefore, they are electrolytes.

In solid state ions are arranged into crystal lattice and will not conduct a current of electricity.

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What is the density of a 10 kg mass that occupies 5 liters?<br> ( pls need help)
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Answer: d=2000 g/L

Explanation:

Density is mass/volume. The units are g/L. Since we are given mass and volume, we can divide them to find density. First, we need to convert kg to g.

10kg*\frac{1000g}{1kg} =10000 g

Now that we have grams, we can divide to get density.

d=\frac{10000g}{5 L}

d=2000g/L

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When ethane (C2H6) burns, it produces carbon dioxide and water:
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The force needed to make an object with a mass of 5kg accelerate at 10 m/sec2 = Question 5 options:
astraxan [27]

Explanation:

<h3>Here, mass = 5kg </h3><h3> acceleration = 10m/s^2</h3>

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8 0
3 years ago
The following data was collected for the formation of ammonia (NH3) based on the following overall reaction: N2 + 3H2 = 2NH3 N2
notsponge [240]

Answer :  The unit for the rate constant in the rate law for the formation of ammonia is, M^{-2}min^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

N_2+3H_2\rightarrow 2NH_3

Rate law expression for the reaction:

\text{Rate}=k[N_2]^a[H_2]^b

where,

a = order with respect to N_2

b = order with respect to H_2

Expression for rate law for first observation:

0.0021=k(0.10)^a(0.10)^b ....(1)

Expression for rate law for second observation:

0.0084=k(0.10)^a(0.20)^b ....(2)

Expression for rate law for third observation:

0.0672=k(0.20)^a(0.40)^b ....(3)

Dividing 2 by 1, we get:

\frac{0.0084}{0.0021}=\frac{k(0.10)^a(0.20)^b}{k(0.10)^a(0.10)^b}\\\\4=2^b\\b=2

Dividing 3 by 1 and also put value of b, we get:

\frac{0.0672}{0.0021}=\frac{k(0.20)^a(0.40)^2}{k(0.10)^a(0.10)^2}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[N_2]^a[H_2]^b

\text{Rate}=k[N_2]^1[H_2]^2

Now, calculating the value of 'k' by using any expression.

0.0021=k(0.10)^1(0.10)^2

k=2.1M^{-2}min^{-1}

The value of the rate constant 'k' for this reaction is 2.1M^{-2}min^{-1}

That means, the unit for the rate constant in the rate law for the formation of ammonia is, M^{-2}min^{-1}

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I’m am most positive the answer is c. For cat
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