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3 years ago

Which of the following substances contains a nonpolar covalent bond?

Chemistry

1 answer:

garri49 [273]3 years ago

6 0

B) H₂

Explanation:

Only H₂ is a non-polar covalent molecule among the given options. A non-polar covalent bond forms between homonuclear atoms that are chemically bonded.

In the formation of covalent bonds, each atom shares the valence electrons in their outer energy shells to be able to mimic the noble gases.
In homonuclear molecules, the two bonding atoms are of the same kind and the electronegativity difference is zero.
The covalent bond between them is a called a non-polar covalent bond.
Here there is equal sharing of the electron pair between the pair of atoms and there is no electrostatic charge on the molecule.

Learn more:

chemical bonds brainly.com/question/10903097

carbon bonds brainly.com/question/10901976

bond types brainly.com/question/11730855

#learnwithBrainly

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Do equal volumes of the same gas at the same temperature and pressure contain equal numbers of molecules (number of moles)?

tresset_1 [31]

Answer:Therefore, there must be a direct relationship between these volumes of gases and the number of molecules they contain. Avogadro's law says that: Equal volumes of different gaseous substances, measured under the same pressure and temperature conditions, contain the same number of molecules.

Explanation:I hope it works for you.

4 0

3 years ago

The molar heat of vaporization of water is 40.7kJ/mol. How much heat must be absorbed to convert 50.0 grams of liquid water at 1

EleoNora [17]

During a phase change the temperature does not change since all of the heat is being absorbed in order to break the intermolecular forces. Due to that, the formula will not need to have T in it and is actually q=nΔH(v).
n=the number of moles (in this case 2.778mol of water since you divide 50g by 18g/mol).
ΔH(v)=the molar heat of vaporization (in this case 40.7kJ/mol).
q=the heat that must be absorbed
q=2.778mol×40.7kJ/mol
q=113.1kJ
Therefore the water needs to absorb 1.13×10²kJ.

I hope this helps. Let me know if anything is unclear.

4 0

3 years ago

What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?

kari74 [83]

The heat of reaction (i.e. combustion) of butane ( $C_{4} H_{10}$ ) when reacted with oxygen ( $O_{2}$ )  is -2658 kJ/mol butane, and the chemical reaction is given by:

$C_{4} H_{10}$ + $\frac{13}{2} O_{2}$ ---> $4 CO_{2}$ + $5 H_{2}O$

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants.

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required.

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane = 32.73 grams butane

The mass of carbon dioxide ( $CO_{2}$ ) can be determined by multiplying the moles of butane ( $C_{4} H_{10}$ ) with the mole ratio of ( $CO_{2}$ ) produced to the ( $C_{4} H_{10}$ ) reacted, and then with the molar mass of ( $CO_{2}$ ), which is 44 grams/mole.

Mass of carbon dioxide produced
= 0.5643 moles butane * [4 moles $CO_{2}$ / 1 mole $C_{4} H_{10}$ ] * 44 grams/mole $CO_{2}$

Mass of carbon dioxide produced
= 99.32 grams $CO_{2}$

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.

4 0

3 years ago

Read 2 more answers

The average rate of disappearance of ozone in the reaction 2o3(g) → 3o2(g) is found to be 7.25×10–3 atm over a certain interval

worty [1.4K]

<h3><u>Answer</u>;</h3>

1.0875 x 10-2 atm

<h3><u>Explanation;</u></h3>

2O3(g) → 3O2(g)

rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t

The average rate of disappearance of ozone ... is found to

be 7.25 × 10–3 atm over a certain interval of time.

This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm

(it is disappearing, thus the negative sign)

rate = -(1/2)∆[O3]/∆t

rate = -(1/2)*(-7.25 × 10^–3 atm)

= 3.625 × 10^–3 atm

Now use the other part of the expression:

rate = +(1/3)∆[O2)∆t

3.625 × 10–3 atm = +(1/3)∆[O2)/t

∆[O2)/∆t = (3)*(3.625× 10^–3 atm)

= 1.0875 x 10-2 atm over the same time interval

4 0

3 years ago

HELP ASAP!! PLEASE !!

ZanzabumX [31]

Answer:

8,3, 7,7

Explanation:

6 0

2 years ago