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vampirchik [111]
11 months ago
11

What mass of H atoms is contained in 50.7 g of NH3? Answer in units of g.

Chemistry
1 answer:
Ratling [72]11 months ago
7 0

The mass of hydrogen present in 50.7 grams of ammonia is equal to 8.95 g.

<h3>What is a mole?</h3>

A mole is a scientific unit that is used to determine the huge number of quantities of atoms, molecules, ions, etc. The mass of the 1 mole of any element is called atomic mass and that of one mole of any compound is called molar mass.

The number of entities present in 1 mole was found to be equal to 6.023 × 10 ²³ per mole which is known as Avogadro’s constant.

Given, the mass of ammonia = 50.7 g

The mass of the one mole of ammonia = 17 g/mol

17 g of ammonia contain the mass of hydrogen = 3 g

50.7 g of ammonia contains a mass of hydrogen = (3/17) × 50.7 = 8.95 g

Learn more about the mole, here:

brainly.com/question/26416088

#SPJ1

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Draw the product formed when cyclohexene is reacted with H2 in the presence of Pt. Note: If adding hydrogen atoms to a carbon at
tigry1 [53]

Answer:

It has been drawn and uploaded as an attachment. Please download it to see the structure.

Explanation:

The product formed as a result of the reaction of cyclohexene with H2​ in presence of Pt (platinum) can be described as catalytic hydrogenation. Catalytic hydrogenation is defined as the process of hydrogen addition in the presence of a catalyst, which in this case is platinum.

Note that Cyclohexene (alkene) is a hydrocarbon molecule represented by the chemical formula, C6​H10​ .

It consists of a double bond. During the hydrogenation reaction, the alkene undergoes an addition reaction to give alkane which is a saturated hydrocarbon as the product.

The first step in order to derive the product is to draw the chemical structure of cyclohexene and identify the double bond present in it.

The final product can be derived by replacing the double bond with the single bond and satisfying all the valences of the carbon atom. The final product structure has been drawn and uploaded as an attachment. Please download it to see the structure.

Ans:

The structure of the cyclohexane thus, formed has been shown as follows with all the hydrogen atoms:

3 0
3 years ago
Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of
Brut [27]

<u>Answer:</u> The smallest temperature change is shown by water.

<u>Explanation:</u>

To calculate the heat absorbed or released, we use the equation:

q=mC\times \Delta T      ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

\Delta T = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

  • <u>Option a:</u>  50.0 g Fe, C_{Fe}=0.449J/g^oC

We are given:

m=50.0g\\C_{Fe}=0.449J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC

Change in temperature = 8.99°C

  • <u>Option b:</u>  50.0 g water, C_{water}=4.18J/g^oC

We are given:

m=50.0g\\C_{water}=4.18J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC

Change in temperature = 0.96°C

  • <u>Option b:</u>  25.0 g Pb, C_{Pb}=0.128J/g^oC

We are given:

m=50.0g\\C_{Pb}=0.128J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC

Change in temperature = 62.5°C

  • <u>Option d:</u>  25.0 g Ag, C_{Ag}=0.235J/g^oC

We are given:

m=25.0g\\C_{Ag}=0.235J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC

Change in temperature = 34.04°C

  • <u>Option e:</u>  25.0 g granite, C_{granite}=0.79J/g^oC

We are given:

m=25.0g\\C_{Fe}=0.79J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

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