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Hunter-Best [27]
2 years ago
10

A 0.100 M solution of K2SO4 would contain the same total ion concentration as which of the following solutions?0.0800 M Na2CO3 0

.100 M NaCl 0.0750 M Na3PO4 0.0500 M NaOH
Chemistry
1 answer:
AysviL [449]2 years ago
7 0

Answer:

  • <em><u>The third chice: 0.0750 M Na₂SO₄</u></em>

Explanation:

Assume 100% ionization:

<em><u>1) 0.100 M solution K₂SO₄</u></em>

  • K₂SO4 (aq) → 2K⁺ (aq) + SO₄²⁻ (aq)

  • Mole ratios: 1 mol K₂SO4 : 2 mol K⁺  + 1 mol SO₄²⁻ (aq) : 3 mol ions. This is 1 : 3

  • At constant volume, the mole ratios are equal to the concentration ratios (M).

  • 1  M K₂SO₄: 3 M ions = 0.100 M K₂SO₄ / x ⇒ x = 0.300 M ions

This means, that you have to find which of the choices is a solution that contains the same 0.300 M ion concentration.

<u>2) 0.0800 M Na₂CO₃</u>

  • Na₂CO₃ (aq) → 2 Na⁺ + CO₃⁻

  • 1 M Na₂CO₃ / 3 M ions = 0.0800M / x ⇒ x = 0.0267 M ions

This is not equal to 0.300 M, so this solution would not contain the same total concentration as a 0.100 M solution of K₂SO₄, and is not the right answer.

<u>3)  0.100 M NaCl </u>

  • NaCl → Na⁺ + Cl⁻

  • 1 M NaCl / 2 M ions = 0.100 M NaCl / x ⇒ x = 0.200 M ions

This is not equal to 0.300 M ion, so not a correct option.

<u>4) 0.0750 M Na₃PO₄</u>

  • Na₃PO₄ → 3Na⁺ + PO₄³⁻

  • 1 M Na₃PO₄ / 4 M ions = 0.0750 M Na₃PO₄ / x ⇒ x = 0.300 M ions

Hence, this ion concentration is equal to the ion concentration of a 0.100 M solution of K₂SO₄, and is the correct choice.

<u>5)  0.0500 M NaOH </u>

  • NaOH → Na⁺ + OH⁻

  • 1 M NaOH / 2 mol ions = 0.0500 M NaOH / x ⇒ x = 0.100 M ions

Not equal to 0.300 M, so wrong choice.

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weeeeeb [17]

Answer : The grams of N_2 consumed is, 89.6 grams.

Solution : Given,

Mass of CaCN_2 = 265 g

Molar mass of CaCN_2 = 80 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate the moles of CaCN_2.

\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles

The given balanced reaction is,

CaC_2+N_2\rightarrow CaCN_2+C

from the reaction, we conclude that

As, 1 mole of CaCN_2 produces from 1 mole of N_2

So, 3.2 moles of CaCN_2 produces from 3.2 moles of N_2

Now we have to calculate the mass of N_2

\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2

\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g

Therefore, the grams of N_2 consumed is, 89.6 grams.

5 0
2 years ago
Read 2 more answers
A penny has a mass of
Ahat [919]
You did not include the questions.

I did some research and found the questions:

<span> What is the mass of 1 mole of pennies? How many moles of pennies have a mass equal to the mass of the moon?

Solutions:

1) mass of 1 mole of pennies

Data: mass of 1 penny = 2.50 g

1 mole = 6.022 * 10^ 23 units

Proportion:

  1 penny      6.022 * 10^23 penny
-------------- = ----------------------------
   2.50 g                    x

Solve: x = 6.022 * 10^23 penny * 2.50g / 1 penny = 15.055* 10^23

Since 2.50 has 3 significant figures, the answer must use 3 significant figures => x = 15.1 * 10^ 23 g = 1.51 * 10^24 g

Answer: 1 mol of pennies have a mass of 1.51 * 10^24 g

2) How many moles of pennies have a mass equal to the same mass of the Moon

Convert the mass of the Moon grams: 7.35 * 10^22 kg = 7.35 * 10^ 25 g

       1 mol                            x
---------------------- =  ----------------------
1.51 * 10^ 24g          7.35 * 10^ 25 g

=> x = 7.35 * 10^ 25 g * 1 mol / (1.51 * 10^24 g)= 48.7 mol

Answer: 48.7 mol
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5 0
2 years ago
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What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2?
Anna11 [10]
Molar mass  CaCl₂ =  110.98 g/mol

Number of moles:

1 mole CaCl₂ ---------> 110.98 g
n mole CaCl2 ---------> 85.3 g

n = 85.3 / 110.98

n = 0.7686 moles of CaCl₂

Volume = ?

M = n / V

0.788 =  0.7686 / V

V = 0.7686 / 0.788

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hope this helps!
5 0
2 years ago
Which of the following reactions would have the smallest value of K at 298 K? Which of the following reactions would have the sm
Firdavs [7]

Answer:

The reaction with smallest value of K is :

A + B → 2 C; E°cell = -0.030 V

Explanation:

nFE^o_{cell}=RT\ln K

where :

n = number of electrons transferred

F = Faraday's constant = 96500 C

E^o_{cell} = standard electrode potential of the cell

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = 25^oC=[273+25]=298K

K = equilibrium constant of the reaction

As we cans see, that standard electrode potential of the cell is directly linked to the equilibrium constant of the reaction.

  • Higher E^o_{cell} higher will be the value of K.
  • Lower E^o_{cell} lower will be the value of K.

So, the reaction with smallest value of electrode potential will have smallest value of equilibrium constant. And that reaction is:

A + B → 2 C; E^o_{cell} =-0.030 V

7 0
2 years ago
Many metals pack in cubic unit cells. The density of a metal and length of the unit cell can be used to determine the type for p
jarptica [38.1K]

Answer:

A.) 34.866 x 10¯23 g/atom

B.) 2.64 x 10^22 atoms in 1 cm3

C.) 1 atoms per unit cell

Explanation:

A) Calculate the average mass of one atom of polonium:

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B) Determine atoms in 1 cm3:

9.19g / 34.866x 10¯23 g/atom = 2.64 x 10^22 atoms in 1 cm3

Determine volume of the unit cell:

(3.37 x 10¯8 cm)3 = 3.827 x 10¯23 cm3

Determine number of unit cells in 1 cm3:

1 cm3 / 3.827 x 10¯23 cm3 = 2.61 x 1022 unit cells

C) Determine atoms per unit cell:

2.64 x 1022 atoms / 2.61 x 1022 unit cells = 1 atoms per unit cell

3 0
2 years ago
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