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valkas [14]
3 years ago
11

HELP ME PLZZ I NEED HLELP!! PLEASE DONT GIVE A WRONG ANSWER

Chemistry
1 answer:
Schach [20]3 years ago
7 0

Answer: B

Explanation:

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Calculate the volume in milliliters of a 1.3M zinc nitrate solution that contains 100.g of zinc nitrate ZnNO32. Be sure your ans
zheka24 [161]

Answer:

The volume is 406 mL

Explanation:

Step 1: Data given

Molarity of a zinc nitrate solution = 1.3 M

Mass of zinc nitrate = 100 grams

Molar mass of Zn(NO3)2 = 189.36 g/mol

Step 2: Calculate moles Zn(NO3)2

Moles Zn(NO3)2 = mass Zn(NO3)2 / molar mass Zn(NO3)2

Moles Zn(NO3)2 = 100.0 grams / 189.36 g/mol

Moles Zn(NO3)2 = 0.5281 moles

Step 3: Calculate volume

Molarity = moles / volume

Volume = moles / molarity

Volume = 0.5281 moles / 1.3 M

Volume = 0.406 L = 406 mL

The volume is 406 mL

3 0
4 years ago
Help ASAP! I'm kind of stuck on what you are supposed to put for theoretical / predicted yield and what is the actual yield for
Alex777 [14]
The theoretical yield of a product is the calculated amount of what you are supposed to get by the end of a reaction assuming that nothing is lost, but the actual yield of the product is what you actually get when you carry out the reaction in real life conditions.
3 0
3 years ago
Fission reactions can get out of control without what taking place?Select one of the options below as your answer:A. chain react
guajiro [1.7K]
Fission reactions can get out of control without neutron moderation. The correct option among all the options that are given in the question is the fourth option or option "D". The other options in this question can be discarded. I hope that this is the answer that you were looking for and it has come to your help.
3 0
3 years ago
Read 2 more answers
Write the balanced half-equation describing the oxidation of mercury to hgo in a basic aqueous solution. Please include the stat
Cloud [144]

Answer:

Hg^0+2OH^-\rightarrow Hg^{2+}O+H_2O+2e^-

Explanation:

Hello,

In this case, mercury (II) oxide (HgO) is obtained via the reaction:

Hg(l)+O_2\rightarrow HgO

Nonetheless, since it is a reaction carried out in basic solution, mercury's half-reaction only, must be:

Hg^0+2OH^-\rightarrow Hg^{2+}O+H_2O+2e^-

Thus, it is seen that OH ionis should be added due to the basic aqueous solution considering that 2 electrons are transferred from 0 to 2 in mercury.

Best regards.

8 0
3 years ago
If the reaction of 150. G of ammonia with 150. G of oxygen gas yields 87. G of nitric oxide (no), what is the percent yield of t
Gekata [30.6K]

The percentage yield of the given reaction is 77.33%.

What is percentage yield?
Reactants
frequently produce fewer product quantities than predicted by the chemical reaction's formula. The percentage of a theoretical yield that's been produced in a reaction is calculated using the percent yield formula. Working through a stoichiometry problem yields the theoretical yield, which is the ideal quantity of the final product. The actual yield is determined by calculating the volume of the product formed. We can determine the percentage yield by dividing the actual yield by the theoretical yield.

Moles is calculated by using the formula:
Moles of Ammonia:
Given mass of ammonia = 150g
Molar mass of ammonia = 17 g/mol
Putting values in above equation, we get:

Moles of Oxygen
Given mass of oxygen = 150g
Molar mass of oxygen = 32 g/mol
Putting values in above equation, we get:
For the given chemical equation:

By Stoichiometry,
5 moles of oxygen reacts with 4 moles of ammonia.
So, 4.6875 moles of oxygen will react with =  of ammonia
As, moles of ammonia required is less than the calculated moles. Hence, ammonia is present in excess and is termed as excess reagent.
Therefore, oxygen is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the given reaction:
5 moles of oxygen gas produces 4 moles of nitric oxide
So, 4.6875 moles of oxygen gas will produce =  of nitric oxide

Now, to calculate the theoretical amount of nitric oxide, we use equation 1: Molar mass of nitric oxide = 30 g/mol
 Given mass of nitric oxide = 112.5 g
Now, to calculate the percentage yield, we use the formula:
Experimental yield = 87 g
Theoretical yield = 112.5 g
Putting values in above equation, we get:
Hence, the percentage yield of the given reaction is 77.33%.

learn more about percentage yield
brainly.com/question/14714924
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6 0
1 year ago
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