Name the following compound: CH₂-CH-CH₂ - CH₂ - CH3 1 2 CH₂ CH₂

compute the mass of caso4 that can be prepared by the reaction of 3.2900g of h2so4 with 3.1660g of caco3

viva [34]

1 mole of calcium carbonate reacts with 1 mole of sulfuric acid and produces 1 mole of calcium sulfate. 3.1660 g of CaCO3 is how many moles of calcium carbonate? 3.1660 / 100.0869 = 0.031633 moles. 3.2900 g of H2S04 is how many moles of sulfuric acid? 3.2900 / 98.079 = 0.033544 moles.
The lesser of the two is 0.031633 moles. Therefore, 0.031633 moles of calcium carbonate will combine with 0.031633 moles of sulfuric acid to produce 0.031633 moles of calcium sulfate. Molecular weight of calcium sulfate is 136.14 g/mol. Therefore, 0.031633 moles of calcium sulfate will weight 0.031633 x 136.14 g/mol = 4.3065 grams.

5 0

3 years ago

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What is meant by isomerism?

agasfer [191]

What is meant by isomerism?

➪ Two or more compounds having the same molecular formula but different structures are called isomers and this process is known as isomerism...~

8 0

2 years ago

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In an effort to try winning World War I, many countries broke previously created treaties and pacts that banned the use of chemi

nadya68 [22]

D the belguim people atacked using it

3 0

3 years ago

The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is belie

Deffense [45]

Answer: Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

Explanation:

To calculate the number of moles, we use the equation

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

For ozone:

Given mass of ozone = 0.827 g

Molar mass of ozone = 48 g/mol

Putting values in above equation, we get:

$\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol$

For nitric oxide:

Given mass of nitric oxide = 0.635 g

Molar mass of nitric oxide = 30.01 g/mol

Putting values in above equation, we get:

$\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol$

For the given chemical equation:

$O_3+NO\rightarrow O_2+NO_2$

By Stoichiometry of the reaction:

1 mole of ozone reacts with 1 mole of nitric oxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitric oxide

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

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

Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles

By Stoichiometry of the reaction:

1 mole of ozone produces 1 mole of nitrogen dioxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitrogen dioxide

Now, calculating the mass of nitrogen dioxide from equation 1, we get:

Molar mass of nitrogen dioxide = 46 g/mol

Moles of nitrogen dioxide = 0.0172 moles

Putting values in equation 1, we get:

$0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g$

Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

8 0

3 years ago

What is the answer? show proof I will report you if you answer for the points

IceJOKER [234]

Answer:

A

Explanation:

carbon dioxide is released in cellular respiration

3 0

3 years ago