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dmitriy555 [2]
2 years ago
14

Solid aluminum metal and diatomic chlorine gas react spontaneously to form a solid product. Give the balanced chemical equation

(including phases) that describes this reaction.
Chemistry
1 answer:
ValentinkaMS [17]2 years ago
8 0

When solid aluminum metal is reacted with diatomic chlorine gas, solid aluminum chloride is formed. This reaction is an example of synthesis or chemical combination in which two elements, aluminum and chlorine combine to form a new compound aluminum chloride.

Word equation: Aluminum (s)+ Chlorine (g)---> Aluminum chloride(s)

Molecular formula of the product formed is AlCl_{3}.

Therefore the balanced chemical equation representing the reaction of solid aluminum with gaseous dichlorine can be represented as,

2Al(s) + 3Cl_{2}(g)-->2AlCl_{3}(s)

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A sealed vessel contains 0.200 mol of oxygen gas, 0.100 mol of nitrogen gas, and 0.200 mol of argon gas. The total pressure of t
babunello [35]

Answer:

D

Explanation:

We can use the mole ratio to calculate the partial pressure. The total number of moles is 0.2 + 0.2 + 0.1 = 0.5 moles

Now, we know that the mole fraction of the argon gas would be 0.2/0.5

The partial pressure is as follows. To calculate this, we simple multiply the number of moles by the total pressure.

0.2/0.5 * 5 = 1.0/0.5 = 2.00atm

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2 years ago
If 14c-labeled uracil is added to the growth medium of cells, what macromolecules will be labeled?
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When there are 14c-lable uracil that are added to the growth medium of cells, the macromolecules that will be labled are RNA. Uracil is a nucleobase that make up the DNA or the RNA. In RNA, uracil binds with other nucleobase (adenine) through hydrogen bonds.
6 0
2 years ago
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Acetic acid is a weak acid with a pKa of 4.76. What is the concentration of acetate in a buffer solution of 0.2M at pH 4.9. Give
Ghella [55]

Answer:

[base]=0.28M

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:

pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\

[base]=1.38[acid]=1.38*0.20M=0.28M

Regards.

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