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fomenos
3 years ago
13

Which formula can be used to calculate the molar mass of ammonia

Chemistry
1 answer:
Nikolay [14]3 years ago
7 0

Answer:

Since there are 3 Hydrogen atoms present, the formula mass of H is 1.0 × 3 = 3.0 g/mol. Therefore, by adding them up, the formula mass of ammonia is: [14.0 g/mol + 3.0 g/mol] = 17.0 g/mol.

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How many different values of ml are possible in the 2p sublevel?
Kisachek [45]

There are four quantum numbers for an orbital, namely, principal quantum number n, azimuthal quantum number l, magnetic quantum number m_{l} and spin quantum number s. No two orbitals can have same value for all the four quantum numbers.

Principal quantum number tells about the position of orbital that is shell number, azimuthal quantum number describes the shape of orbital, for s, p, d, f,... orbitals, the value is 0, 1, 2, 3,... so on. The value of magnetic quantum number varies from -l to +l (including zero), it defines the sub shell of electron. Spin of the electron defines its movement that is clockwise or anticlockwise thus, there are only two possible values for spin quantum number: +1/2 or -1/2.

For 2p orbital, principal quantum number n is 2 (denotes the shell) and azimuthal quantum number for p orbital is 1. Thus, different values for magnetic quantum number will be -l to l that is -1, 0,1.

Therefore, for 2p subshell there are 3 different  m_{l} values.

5 0
3 years ago
Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.
weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

OR

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

7 0
3 years ago
Three dimensional scale model of the Earth
Likurg_2 [28]
A globe , a sphere could be used as a three dimensional mode of the earth as well.
6 0
3 years ago
Consider the following balanced equation:
kow [346]

Answer:

The theoretical yield of Cu(s) in moles is 60.15 moles

Explanation:

Step 1: Data given

Number of moles CuO = 70.8 moles

Number of moles NH3 = 40.1 moles

Molar mass CuO = 79.545 g/mol

Molar mass NH3 = 17.03 g/mol

Step 2: The balanced equation

3CuO(s) + 2NH3(g) → 3H2O(l) + 3Cu(s) + N2(g)

For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2

NH3 is the limiting reactant. It will completely be consumed (40.1 moles). CuO is in excess. There will react 3/2 * 40.1 = 60.15 moles

There will remain 70.8 - 60.15 = 10.65 moles CuO

Step 3: Calculate moles Cu

For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2

For 40.1 moles NH3 we'll have 60.15 moles Cu

The theoretical yield of Cu(s) in moles is 60.15 moles

8 0
4 years ago
What biological macromolecule is made up of monomers like the one shown below? NH2-RHC - COOH. A. Lipid B. Nucleic Acid C. Prote
ololo11 [35]
The biological macromolecule that is made of those monomers is C. Protein.
3 0
3 years ago
Read 2 more answers
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