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3 years ago

Which mineral test was used if a geologist refers to a mineral as metallic?

Chemistry

2 answers:

Harlamova29_29 [7]3 years ago

5 0

Answer: Luster

Explanation:

A mineral is a natural solid that has a a definite chemical composition.

There are two kinds of minerals : metallic and non -metallic

The mineral test was used if a geologist refers to a mineral as metallic is luster because luster gives a way in which a metal reflects the light.

If a mineral is metallic then it will reflect light.

Hence, Luster was used if a geologist refers to a mineral as metallic.

Stolb23 [73]3 years ago

4 0

Streak. It couldn't be hardness because some minerals are soft and metallic, and some are hard and metallic. Also, it can't be luster, because most gems are minerals, and most luster, but they're not metallic. And it can't be magnetism, because magnetite, the only magnetic mineral, is not really metallic.

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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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A gram of gasoline produces 45 kJ of energy when burned. Gasoline has a density of 0.77 g/mL. How would you calculate the amount

BigorU [14]

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borishaifa [10]

Answer:

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