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3 years ago

15

Which of the following can form a hydrogen bond with the HF molecule?

1 answer:

Lostsunrise [7]3 years ago

6 0

Answer:

helium

(He)

Explanation:

helium

(He)

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You do not know an object's velocity until you know its what?

luda_lava [24]

You do not know an object's velocity until you know its speed and direction

6 0

4 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

4 years ago

When the problem within the story gets more intense and builds, this is the... Select one: climax. falling action. rising action

timama [110]

Answer:

rising action.

Explanation:

Rising action is a term used in literature to describe the corresponding portion of the story, that occurs immediately before the climax, by increasing its atmosphere or builds-up whereby the tension level surges.

Climax, on the other hand, is a form of the highest point major turning point in a story.

While falling action is the point where there is a quick decrease in tension level, and follows immediately after the climax

Resolution is also known as the final part of the story, it follows immediately after falling action.

Hence, when the problem within the story gets more intense and builds, this is the "Rising Action."

4 0

3 years ago

A stream leaving a mountain range deposits a large part of its load in a(n): alluvial fan sandbar meander sand dune

pshichka [43]

the answer to you question is alluvial fan

3 0

3 years ago

Read 2 more answers

One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, which is a greenhouse gas; that i

vagabundo [1.1K]

Answer:

The value is $U =1.3908 *10^{11} \ kg$

Explanation:

From the question we are told that

mass of carbon dioxide produced by one gallon of gasoline is $m = 9.50 \ kg$

The number of cars is $N = 40 \ million = 40 *10^{6} \ cars$

The distance covered by each car is $d = 7930 \ mi$

The rate is $R = 23.6 \ mi/ gallon$

Generally the amount of gasoline used by one car is mathematically represented as

$G = \frac{d}{R}$

=> $G = \frac{7930}{23.6}$

=> $G = 366 \ gallons$

Generally the amount of gasoline used by N cars is

$H = N * G$

=> $H = 40*10^{6} * 366$

=> $H = 1.464*10^{10} \ gallons$

Generally the annual production of carbon dioxide is mathematically represented as

$U = m * H$

=> $U =9.50 * 1.464*10^{10}$

=> $U =1.3908 *10^{11} \ kg$

8 0

4 years ago