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Mariulka [41]
3 years ago
6

Beryllium is a rare metal that is gray in color, strong, and lightweight. The following data was obtained from a calorimeter exp

eriment: 5 sets of equations: m beryllium equals 10.1 grams. M water equals 38.8 grams. Delta T beryllium equals final T minus initial T equals 27.7 degrees Celsius minus 100 degrees Celsius. Delta T water equals Final T minus initial T equals 28.4 degrees Celsius minus 20.2 degrees Celsius. C water equals 4.18 joules per gram per degree Celsius. The specific heat of beryllium is J/g°C
Chemistry
2 answers:
arsen [322]3 years ago
8 0

Answer:

Explanation:

Correct Answer is : 1.82 J/g°C

Elanso [62]3 years ago
5 0

Answer:  1.82 J/g°C

Explanation:

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<u>Answer:</u> 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.

<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 ozone:</u>

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

  • <u>For nitric oxide:</u>

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.

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