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3 years ago

Alkynes have what in common with alkenes ?

Chemistry

1 answer:

jonny [76]3 years ago

8 0

Answer:

Alkanes, alkenes and alkynes are simple hydrocarbon chains with no functional groups. Alkanes are identified because the carbon chain has only single bonds. Alkenes have at least one double bond and alkynes have at least one triple bond. The most common alkyne is ethyne, better known as acetylene.

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Why fluorine is more electronegative than both oxygen and chlorine.

ziro4ka [17]

Why chlorine has highest electron affinity than fluorine?

This is because the atomic radius increases down a group. The electron gained ends up in the outermost shell. ... Fluorine, which is higher up the group then chlorine, has a lower electron affinity. This is because the electrons in the outermost shell of a fluorine atom are closer together.

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2 years ago

An OH group attached to a hydrocarbon is called a _________ group whereas ______________ is a polyatomic ion with a charge of __

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3 years ago

Calculate the volume occupied by 272g

cupoosta [38]

The answer for the following problem is mentioned below.

Therefore volume occupied by methane gas is 184.78 × 10^-3 liters

Explanation:

Given:

mass of methane( $CH_{4}$ ) = 272 grams

pressure (P) = 250 k Pa =250×10^3 Pa

temperature(t) = 54°C =54 + 273 = 327 K

Also given:

R = 8.31JK-1 mol-1 ,

Molar mass of methane( $CH_{4}$ ) = 16.0 grams

We know;

According to the ideal gas equation,

P × V = n × R × T

here,

n = m÷M

n =272 ÷ 16

n = 17 moles

Therefore,

250×10^3 × V = 17 × 8.31 × 327

V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )

V = 184.78 × 10^-3 liters

Therefore volume occupied by methane gas is 184.78 × 10^-3 liters

6 0

3 years ago

What is the de Broglie wavelength (in meters) of a 45-g golf ball traveling at 72 m/s?

Cerrena [4.2K]

Answer: The de broglie wavelength is $2.037 \times 10^{-34} m$ .

Explanation:

Calculate $\lambda = \frac{h}{p}$ as follows.

$\lambda = \frac{h}{p}$

where,

h = plank's constant = $6.6 \times 10^{-34} m^{2} kg/s$

p = momentum = $mass \times velocity$

Putting the values in the formula as follows.

$\lambda = \frac{h}{mass \times velocity}$

= $\frac {6.6 \times 10^{-34} m^{2} kg/s}{0.045 kg \times 72 m/s}$

= $2.037 \times 10^{-34} m$

Thus, the de broglie wavelength is $2.037 \times 10^{-34} m$ .

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3 years ago

Read 2 more answers

What is one factor that contributes to seasons occurring in Pennsylvania?

JulijaS [17]

B. the tilt of the Earth on it's axis

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3 years ago