All categories

3 years ago

Name the compound CH3CH double bond C(CH3)CH(CH3)2.

1 answer:

4 0

CH3CH=C(CH3)CH(CH3)2

Start the carbon count at the end closest to the most important element, the double bond. I would recommend drawing the skeletal structure too so you can see where carbons are attached and how long the chain is.

3,4- dimethyl-2-pentene

You might be interested in

Which of the following is true regarding the atoms involved in a chemical reaction?

andrezito [222]

Is there a picture to this question?

4 0

3 years ago

Percent mass composition

goldfiish [28.3K]

Answer:Mass percent composition is also known percent by weight. It is abbreviated as w/w%. For a solution, mass percent equals the mass of an element in one mole of the compound divided by the molar mass of the compound, multiplied by 100%.

3 0

3 years ago

Please help me thanks so much?!?.:;)...I’ll mark you.:;)))

oee [108]

A physical change. When a physical change occurs the chemical composition of the substance doesn't change. Some examples of physical changes are melting, freezing, vaporization, and condensation.

5 0

3 years ago

Read 2 more answers

Do it and I will give brainliest

PSYCHO15rus [73]

Answer:

Hi there!

Your answer is

Mass: 12

Electrons:5

Explanation:

The electrons in a neutral atom is the same as the # of Protons!

5 0

3 years ago

A mixture of carbon dioxide and hydrogen gases is maintained in a 6.68 L flask at a pressure of 2.14 atm and a temperature of 19

matrenka [14]

Answer:

The mass of hydrogen gas in the mixture: w₂ = 0.433 g

Explanation:

According to the ideal gas equation:

for an ideal gas, $P.V = n_{total}.R.T$

and $n_{total}= n_{1}+n_{2}$

Here, P: total pressure of the gases = 2.14 atm

V: total volume of the gases = 6.68 L

T: temperature = 19 °C = 19+273.15 = 292.15K (∵ 0°C = 273.15K)

R: gas constant = 0.08206 L·atm·K⁻¹·mol⁻¹

$n_{total}$ : total number of moles of gases

To calculate the total number of moles of gases:

$n_{total} = \frac{P.V}{R.T} = \frac{2.14 atm\times 6.68 L}{0.08206 LatmK^{-}mol^{-}\times 292.15K}$ = 0.5963 moles

Let, the number of moles of carbon dioxide be n₁ and number of moles of hydrogen be n₂

Given: mass of carbon dioxide: w₁ = 16.8 g, mass of hydrogen: w₂ = ?g

molar mass of carbon dioxide: m₁ = 44.01 g/mol, molar mass of hydrogen: m₂= 2.016 g/mol

Therefore, $n_{total}$ = n_{1}+n_{2} = (w₁ ÷ m₁) + (w₂ ÷ m₂)

⇒ 0.5963 mol = (16.8 g ÷ 44.01 g/mol) + (w₂ ÷ 2.016 g/mol)

⇒ 0.5963 mol = (0.3817mol) + (w₂ ÷ 2.016 g/mol)

⇒ 0.5963 mol - 0.3817mol = (w₂ ÷ 2.016 g/mol)

⇒ 0.2146 mol = (w₂ ÷ 2.016 g/mol)

⇒ w₂ = 0.433 g

Therefore, the mass of hydrogen gas in the mixture: w₂ = 0.433 g

4 0

4 years ago

Name the compound CH3CH *double bond* C(CH3)CH(CH3)2.

Name the compound CH3CH double bond C(CH3)CH(CH3)2.