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Anastaziya [24]
2 years ago
6

PLZ HELP! WILL GIVE BRAINLIEST, 5 STARS, AND A THANKS!!

Chemistry
1 answer:
nika2105 [10]2 years ago
8 0

Answer:

The similarities between freezing, and condensation is, the stimulis.

Explanation:

The explanation to that is because the condensation will melt.

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150.0 grams of AsF3 were reacted with 180.0 g of CCl4 to produce AsCl3 and CCl2F2. The theoretical yield of CCl2F2 produced, in
Kamila [148]
The chemical reaction would be written as 

2 AsF3<span> + 3 CCl4 = 2 AsCl3 + 3 CCl2F2
</span>
We use the given amounts of the reactants to first find the limiting reactant. Then use the amount of the limiting reactant to proceed to further calculations.

150 g AsF3 ( 1 mol / 131.92 g) = 1.14 mol AsF3
180 g CCl4 (1 mol / 153.82 g) = 1.17 mol CCl4

 Therefore, the limiting reactant would be CCl4 since it would be consumed completely. The theoretical yield would be:

1.17 mol CCl4 ( 3 mol CCl2F2 / 3 mol CCl4 ) = 1.17 mol CCl2F2

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Water and carbon dioxide into food.
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3 years ago
27.4 g of Aluminum nitrite and 169.9 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams
GarryVolchara [31]

<u>Answer:</u> The mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For aluminium nitrite:</u>

Given mass of aluminium nitrite = 27.4 g

Molar mass of aluminium nitrite = 41 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium nitrite}=\frac{27.4g}{41g/mol}=0.668mol

  • <u>For ammonium chloride:</u>

Given mass of ammonium chloride = 169.9 g

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonium chloride}=\frac{169.9g}{53.5g/mol}=3.176mol

The chemical equation for the reaction of aluminium nitrite and ammonium chloride follows:

Al(NO_2)_3+3NH_4Cl\rightarrow AlCl_3+3N_2+6H_2O

By Stoichiometry of the reaction:

1 mole of aluminium nitrite reacts with 3 moles of ammonium chloride

So, 0.668 moles of aluminium nitrite will react with = \frac{3}{1}\times 0.668=2.004mol of ammonium chloride.

As, given amount of ammonium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium nitrite is considered as a limiting reagent because it limits the formation of product.

Excess moles of ammonium chloride = (3.176 - 2.004) mol = 1.172 moles

Calculating the mass of ammonium chloride by using equation 1, we get:

Excess moles of ammonium chloride = 1.172 moles

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

1.172mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.172mol\times 53.5g/mol)=62.7g

Hence, the mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

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How do you write the formula for potassium hydroxide
makvit [3.9K]

KOH that or H3C-CH2-OH

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List the 3 ways carbon can enter the ocean?
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Carbon dioxide from the atmosphere dissolves in the surface waters of the ocean. Some of the carbon dioxide stays as dissolved gas, but much of it gets turned into other things. Photosynthesis by tiny marine plants (phytoplankton) in the sunlit surface waters turns the carbon into organic matte

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