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ziro4ka [17]
2 years ago
5

Name the 1 isomer of c7h16 that contains an ethyl branch on the parent chain.

Chemistry
1 answer:
cricket20 [7]2 years ago
5 0
An isomer is a compound which has the same atoms as another, but the spatial arrangement of the atoms is different. If we wish to have an ethyl, or two carbon, side chain, then the main chain may be five carbon atoms long. This compound then becomes:

3-ethylpentane; where the parent chain is pentane and the ethyl group is attached to the third carbon from the start of the parent chain.
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Where more of hydrogen are contained in 90 g of water or in 80 g of methane
Lynna [10]
Methane.

Water - H2O
Methane - CH4

Methane has 2 more hydrogens than water.
7 0
3 years ago
____ 26. what is the weight percentage of nitrogen in urea, cn2h4o
Paraphin [41]
Weight percentage of nitrogen can be calculated using the following rule:
weight percentage of nitrogen = (weight of nitrogen / weight of urea) x 100

From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of nitrogen = 14 grams
molecular mass of hydrogen = 1 grams
molecular mass of oxygen = 16 grams

therefore:
mass of nitrogen in urea = 2(14) = 28 grams
mass of urea = 12 + 2(14) + 4(1) + 16 = 60 grams

Substitute with the masses in the equation to get the percentage:
weight percentage of nitrogen = (28/60) x 100 = 46.667%
7 0
3 years ago
Platinum crystallizes in a face-centered cubic cell. the density of platinum is 21.4 g/cm3. calculate the radius of the platinum
IgorLugansk [536]
1 mole of platinum has a mass of 195 g therefore 1 atom will have a mass of 195 g /(6.02 ×10^23) = 3.239 × 10^-22 g
Density is given by dividing mass by volume, thus to get volume, mass is divided by density. 
The volume = (3.239 × 10^-22)/21.4 
                    = 1.514 × 10^-23 cm³
But volume of a sphere is given by 4/3πr³
Therefore, r³ = 3.6129 × 10^-24 
                 r   = ∛(3.6129 × 10^-24)
                      = 1.534 × 10^ -8 cm
Therefore, the radius of the platinum atom is 1.534 × 10^-8 cm

7 0
3 years ago
In the laboratory you are asked to make a 0.175 m barium iodide solution using 13.9 grams of barium iodide. How much water shoul
BARSIC [14]

<u>Answer:</u> The mass of water that should be added in 203.07 grams

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m = molality of barium iodide solution = 0.175 m

m_{solute} = Given mass of solute (barium iodide) = 13.9 g

M_{solute} = Molar mass of solute (barium iodide) = 391.14 g/mol

W_{solvent} = Mass of solvent (water) = ? g

Putting values in above equation, we get:

0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g

Hence, the mass of water that should be added in 203.07 grams

4 0
3 years ago
During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2
ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass  of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol =<u> 12.81 grams</u>

Step 6:  Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = <u>75.70 grams CO2</u>

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = <u>34.29 grams</u>

3 0
3 years ago
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