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3 years ago

Aromatic compounds aliphatic compounds

Chemistry

1 answer:

Nonamiya [84]3 years ago

5 0

Answer:

I hoped it helps you fod blessed:)

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Most elements occur in nature as a single isotope.<br> TRUE<br> FALSE

cestrela7 [59]

Answer:

false

Explanation:

because elements never exist in nature as single isotopes they are always combined

8 0

3 years ago

Read 2 more answers

How many electrons does the copper atom lose in the process: Cu → Cu3+ + ? electrons

alexdok [17]

Answer:

Explanation:

6 0

2 years ago

Two substances A and B, initially at different temperatures, are thermally isolated from their surroundings and allowed to come

Elden [556K]

Answer:

Explanation:

For solving this we need a heat balance

$Q_{a} = Q_{b}\\m_{a}*C_{a}*\Delta T_{a} = m_{b}*C_{b}*\Delta T_{b}$

By changing the corresponding relations, we have

$m_{a}*C_{a}*\Delta T_{a} = \frac{1}{2}m_{a}*4C_{a}*\Delta T_{b} \\\\\\$

By cancelling similar factor, we obtain

$\Delta T_{a} = 2 \Delta T_{b}\\\frac{\Delta T_{a}}{\Delta T_{b}} = 2\\$

Which means that the change of temperature in A is twice the change of B

3 0

3 years ago

The maximum number of electrons in a single d subshell is:

Iteru [2.4K]

10 electrons

Explanation:

The maximum number of electrons in a single d-subshell is 10 electrons.

The d-sub-orbital used to denote azimuthal or secondary quantum numbers.

The maximum number of electrons in the orbitals of sublevels are:

two electrons in the s-sublevel, it has one orbital

six electrons in the p-sublevel, it has three orbital

ten electrons in the d- sublevel, it has five orbitals

fourteen electrons in the f-sublevel, it has seven orbitals

The maximum number of electrons in an orbital is two.

learn more:

Atomic orbitals brainly.com/question/1832385

#learnwithBrainly

4 0

3 years ago

How many joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point?

Korolek [52]

Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.

Amount of heat required to vaporize 1 mole of lead = 177.7 kJ

Molar mass of lead = 207.2 g

Mass of lead given = 1.31 kg = 1310 g (1kg=1000g)

Heat required to vaporize 207.2 of lead = 177.7 kJ

Thus Heat required to vaporize 1310 g of lead = $\frac{177.7}{207.2}\times 1310=1123kJ=1123000J$

Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

7 0

3 years ago