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Serggg [28]
3 years ago
8

In this lab experiment, you will react copper(II) chloride and aluminum. You will look for clues as to which substance was in ex

cess and which limited the reaction. Write an investigative question that describes the scientific goals of this experiment. Make sure that your question applies to any reaction.
Chemistry
2 answers:
timofeeve [1]3 years ago
5 0

<span>The question that applies in this certain scenario in order to satisfy the goal of the experiment is to test the properties of the residues that are formed. These properties may be physical or chemical including the color, reactivity, and other properties that can easily be tested. </span>

GaryK [48]3 years ago
5 0

Answer:

Which is the limiting and excess reactant considering their previously weighted masses?

Explanation:

Hello,

In stoichiometry, the identification of both the limiting and in excess reactants is quite important because based on them one correctly provide the percent yield of a chemical reaction. The limiting reactant is the firstly consumed substance during a chemical reaction and the excess one remain with an unreactive amount, in such a way, all the stoichiometric calculations must be carried out by starting with the limiting reactant's amount (either in grams or moles). Therefore, a suitable question that applies to any reaction, in order to identify both of the aforesaid reactants would be:

Which is the limiting and excess reactant considering their previously weighted masses?

As long as to identify them, we must know the initial reacting masses of both of them.

Best regards.

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A sample of gas in a balloon has an initial temperature of 20. ∘C and a volume of 1.92×103 L . If the temperature changes to 68
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Answer:

\boxed{2.23 \times 10^{3} \text{ L}}

Explanation:

The pressure is constant, so we can use Charles' Law.

\dfrac{ V_{1} }{T_{1}} = \dfrac{ V_{2} }{T_{2}}

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T₂ = (68 + 273.15) K = 341.15 K

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\dfrac{ 1.92 \times 10^{3}}{293.15} = \dfrac{ V_{2}}{341.15}\\\\6.550 = \dfrac{ V_{2}}{341.15}\\\\V_{2} = 6.550 \times 341.15 = 2.23 \times 10^{3} \text{ L}

The new volume of the gas is \boxed{2.23 \times 10^{3} \text{ L}}.

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2 years ago
Determine the density of chunk of lead that has a mass of 1055 grams and a volume of 93.4 cm^3
Bad White [126]
  • Mass=0155g
  • Volume=93.4cm^3

\\ \tt\longmapsto Density=\dfrac{Mass}{Volume}

\\ \tt\longmapsto Density=\dfrac{1055}{93.4}

\\ \tt\longmapsto Density=11.29g/cm^3

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