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3 years ago

Calculate the mass of aluminum that would have the same number of atoms as 6.35 g of cadmium

Chemistry

1 answer:

Sergeu [11.5K]3 years ago

6 0

Answer:

1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.

Explanation:

Given data:

mass of cadmium = 6.35 g

Number of atoms of aluminum as 6.35 g cadmium contain = ?

Solution:

Number of moles of cadmium = 6.35 g/ 112.4 g/mol

Number of moles of cadmium = 0.06 mol

Number of atoms of cadmium:

1 mole = 6.022×10²³ atoms of cadmium

0.06 mol × 6.022×10²³ atoms of cadmium/ 1mol

0.36×10²³ atoms of cadmium

Number of atoms of Al:

Number of atoms of Al = 0.36×10²³ atoms

1 mole = 6.022×10²³ atoms

0.36×10²³ atoms × 1 mol /6.022×10²³ atoms

0.06 moles

Mass of aluminum:

Number of moles = mass/molar mass

0.06 mol = m/ 27 g/mol

m = 0.06 mol ×27 g/mol

m = 1.62 g

Thus, 1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.

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Alborosie

Given:

Concentration of HNO3 = 7.50 M

% dissociation of HNO3 = 33%

To determine:

The Ka of HNO3

Explanation:

Based on the given data

[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M

The dissociation equilibrium is-

HNO3 ↔ H+ + NO3-

I 7.50 0 0

C -2.48 +2.48 +2.48

E 5.02 2.48 2.48

Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23

Ans: Ka for HNO3 = 1.23

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3 years ago

Why don't the praticles in a solid move past one another

Ne4ueva [31]

Answer:

Particles in a solid object are super tight and close to eachother. They do not move past eachother because it is a solid, not a liquid or gas.

Explanation:

3 0

3 years ago

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How many moles of PC15 can be produced from 58.0 g of Cl₂ (and excess P4)?

ludmilkaskok [199]

0.3268 moles of PC15 can be produced from 58.0 g of Cl₂ (and excess

P4)

<h3>How to calculate moles?</h3>

The balanced chemical equation is

$P_{4} + 10Cl_{2} = 4PCl_{5}$

The mass of clorine is m( $Cl_{2}$ ) = 58.0 g

The amount of clorine is n( $Cl_{2}$ ) = m( $Cl_{2}$ )/M( $Cl_{2}$ ) = 58/70.906 = 0.817 mol

The stoichiometric reaction,shows that

10 moles of $Cl_{2}$ yield 4 moles of $PCl_{5}$ ;

0.817 of $Cl_{2}$ yield x moles of $PCl_{5}$

n( $PCl_{5}$ ) = 4*0.817/10 = 0.3268 mol

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brainly.com/question/14935523

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1 year ago

Which aqueous solution has the highest boiling point at standard pressure?(1) 1.0 M KC1(aq) (3) 2.0 M KCl(aq)(2) 1.0 M CaC12(aq)

miss Akunina [59]

Answer:

(4) 2.0 M CaCl₂(aq).

Explanation:

Adding solute to water elevates the boiling point.

The elevation in boiling point (ΔTb) can be calculated using the relation:

ΔTb = i.Kb.m,

where, ΔTb is the elevation in boiling point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kb is the molal elevation constant of water.

m is the molality of the solution.

(1) 1.0 M KCl(aq):

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(1.0 m) = 2(Kb).

(2) 2.0 M KCl(aq):

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(2.0 m) = 4(Kb).

(3) 1.0 M CaCl₂(aq):

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(1.0 m) = 3(Kb).

(4) 2.0 M CaCl₂(aq):

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(2.0 m) = 6(Kb).

So, the aqueous solution has the highest boiling point at standard pressure is: (4) 2.0 M CaCl₂(aq).

6 0

3 years ago

If an ice cube weighing 25.0 g with an initial

riadik2000 [5.3K]

Answer:

∘

Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at

∘

to liquid at

∘

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

−

(

)

, where

- the heat absorbed by the solid at

∘

- the heat absorbed by the liquid at

∘

- the heat lost by the warmer water sample

The two equations that you will use are

⋅

, where

- heat absorbed/lost

- the mass of the sample

- the specific heat of water, equal to

4.18

∘

- the change in temperature, defined as final temperature minus initial temperature

and

⋅

fus

, where

- heat absorbed

- the number of moles of water

fus

- the molar heat of fusion of water, equal to

6.01 kJ/mol

Use water's molar mass to find how many moles of water you have in the

100.0-g

sample

100.0

⋅

1 mole H

18.015

5.551 moles H

So, how much heat is needed to allow the sample to go from solid at

∘

to liquid at

∘

5.551

moles

⋅

6.01

mole

33.36 kJ

This means that equation

(

)

becomes

33.36 kJ

−

The minus sign for

is used because heat lost carries a negative sign.

So, if

is the final temperature of the water, you can say that

33.36 kJ

sample

⋅

sample

−

water

⋅

water

More specifically, you have

33.36 kJ

100.0

⋅

4.18

∘

⋅

(

−

)

∘

−

650

⋅

4.18

∘

⋅

(

−

)

∘

33.36 kJ

418 J

⋅

(

−

)

−

2717 J

⋅

(

−

)

Convert the joules to kilojoules to get

33.36

0.418

⋅

−

2.717

⋅

(

−

)

This is equivalent to

0.418

⋅

2.717

⋅

67.925

−

33.36

34.565

0.418

2.717

11.026

∘

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

∘

Explanation:

3 0

2 years ago

Calculate the mass of aluminum that would have the same number of atoms as 6.35 g of cadmium ​

Calculate the mass of aluminum that would have the same number of atoms as 6.35 g of cadmium