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3 years ago

Do all dipole-dipole forces have the same strength?

1 answer:

n200080 [17]3 years ago

6 0

Inter molecula force
1- hydrogen bond
2- ion bond
3- dipole dipole
4- dipole bond or van der wael
but not all dipole dipole has the same strength

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A current is applied to two electrolytic cells in series. In the first, silver is deposited; in the second, a zinc electrode is

vitfil [10]

The amount of Ag plated out if 1.2 g of Zn dissolves is 3.959 g .

Given ,

A current is applied to two electrolytic cells in series .

In the first ,silver is deposited and in the second a zinc electrode is consumed .

the reactions involving are ;

Ag+ (aq) + e = Ag

Zn = Zn2+ (aq) +2e

thus the resultant equation is ,

2Ag+ (aq) +Zn = 2Ag + Zn2+

Thus for every mole of Zn dissolves , there is 2 moles of Ag is formed .

65.38 g of Zn contains = 1 moles

1.2 g of Zn contains = 1.2/65.38 =0.01835 moles

for every 1 mole of Zn dissolves there is 2 moles of Ag formed .

Thus the amount of Ag formed in moles =2(O.01835) =0.0367 Moles

1 mole of Ag contains = 107.86g

0.0367 moles of Ag contains = 107.86 (0.0367) =3.959 g of Ag

Hence ,the amount of Ag plated out if 1.2 g of Zn dissolves is 3.959 g .

Learn more about electrolytic cell here:

brainly.com/question/19854746

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6 0

1 year ago

Which of the following statements accurately describes the structure of the atom?

Katarina [22]

The answer is b
Protons and neutrons are both in the nucleus
Electrons are outside the nucleus

5 0

3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

Stels [109]

Explanation:

1) Initial mass of the Cesium-137= $N_o$ = 180 mg

Mass of Cesium after time t = N

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0

3 years ago

This is the chemical formula for zinc bromate: . Calculate the mass percent of oxygen in zinc bromate. Round your answer to the

Paladinen [302]

Answer:

30%

Explanation:

<em>This is the chemical formula for zinc bromate: Zn(BrO₃)₂. Calculate the mass percent of oxygen in zinc bromate. Round your answer to the nearest percentage.</em>

Step 1: Determine the mass of 1 mole of Zn(BrO₃)₂

M(Zn(BrO₃)₂) = 1 × M(Zn) + 2 × M(Br) + 6 × M(O)

M(Zn(BrO₃)₂) = 1 × 65.38 g/mol + 2 × 79.90 g/mol + 6 × 16.00 g/mol

M(Zn(BrO₃)₂) = 321.18 g/mol

Step 2: Determine the mass of oxygen in 1 mole of Zn(BrO₃)₂

There are 6 moles of atoms of oxygen in 1 mole of Zn(BrO₃)₂.

6 × m(O) = 6 × 16.00 g = 96.00 g

Step 3: Calculate the mass percent of oxygen in Zn(BrO₃)₂

%O = mO/mZn(BrO₃)₂ × 100%

%O = 96.00 g/321.18 g × 100% ≈ 30%

3 0

3 years ago

The ph of 0.015 m hno2 (nitrous acid) aqueous solution was measured to be 2.63. what is the value of pka of nitrous acid?

Licemer1 [7]

Nitrous acid<span> dissociates as follows:
</span>
HNO₂(s) ⇄ H⁺(aq) + NO₂⁻(aq)

According to the equation, an acid constant has the following form:

Ka = [H⁺] × [NO₂⁻ ] / [HNO₂]

From pH, we can calculate the concentration of H⁺ and NO₂⁻:

[H⁺] = 10^-pH = 10^-2.63 = 0.00234 M = [NO₂⁻]

Now, the acid constant can be calculated:

Ka = 0.00234 x 0.00234 / 0.015 = 3.66 x 10⁻⁴

And finally,

pKa = -log Ka = 3.44

7 0

3 years ago