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3 years ago

The normal boiling point of water is:

Chemistry

1 answer:

Natalka [10]3 years ago

7 0

Answer:

The normal boiling point of water is C) the one measured at 760 mm pressure (1 atm)

Explanation:

The boiling point of water corresponds to 100 ° C (373 K) under pressure conditions of 1 atm (or 760 mmHg or 1013 hPa). It corresponds to the temperature by which the liquid is converted into steam and the point at which vapor pressure is equal to the external atmospheric pressure.

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Which pair of elements is in the same period? (Check ALL that apply) *

stira [4]

Answer:

D and R are in the same period

Explanation:

remember groups are vertical and periods are horizontal!

3 0

2 years ago

Arrange the following compounds in decreasing order of acidity from left to right, and explain the reason in less than 90 words.

saveliy_v [14]

Explanation:

More is the electronegativity of atoms attached to the alpha carbon atom more will be the attraction of electron towards it. As a result, release of hydrogen ion will be easier.

Therefore, there will be increased in acidity of the compound.

Since, fluorine, chlorine and bromine are all group 17 elements and fluorine is more reactive than chlorine. On the other hand, chlorine is more reactive than bromine.

Hence, ability to attract the electrons towards itself will be maximum in fluorine and least in bromine.

Therefore, decreasing order of acidity of the given compounds is as follows.

$CF_{3}COOH$ > $CCl_{3}COOH$ > $CBr_{3}COOH$ > $CH_{3}COOH$

8 0

3 years ago

Estimate ΔH for the reaction: C2H6(g) + Cl2(g)--> C2H5Cl(g) + HCl(g) given the following average bond energies (in kJ/mol): C

Leno4ka [110]

Explanation:

The reaction equation will be as follows.

$C_{2}H_{6}(g) + Cl_{2}(g) \rightarrow C_{2}H_{5}Cl(g) + HCl(g)$

Using bond energies, expression for calculating the value of $\Delta H$ is as follows.

$\Delta H = \sum B.E_{reactants} - \sum B.E_{products}$

On reactant side, from $C_{2}H_{6}$ number of bonds are as follows.

C-C bonds = 1

C-H bonds = 6

From $Cl_{2}$ ; Cl-Cl bonds = 1

On product side, from $C_{2}H_{5}Cl$ number of bonds are as follows.

C-C bonds = 1

C-H bonds = 5

C-Cl bonds = 1

From HCl; H-Cl bonds = 1

Hence, using the bond energies we will calculate the enthalpy of reaction as follows.

$\Delta H = \sum B.E_{reactants} - \sum B.E_{products}$

= $[(1 \times 348 kJ/mol) + (6 \times 414 kJ/mol) + (1 \times 242 kJ/mol)] - [(1 \times 348 kJ/mol) + (5 \times 414 kJ/mol) + (1 \times 327 kJ/mol) + (1 \times 431 kJ/mol)]$ = -102 kJ/mol