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olga55 [171]
2 years ago
11

A chemist has to prepare 250.0 mL of a 0.300 M Na2SO4(aq) solution. What mass, in grams, of sodium sulfate (formula mass 142.05

g/mol) would they use
Chemistry
1 answer:
Sedbober [7]2 years ago
6 0

The mass of sodium sulphate, Na₂SO₄, required to prepare the solution is 10.65 g

<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
  • Volume = 250 mL = 250 / 1000 = 0.25 L
  • Molarity = 0.3 M
  • Mole of Na₂SO₄ =?

Mole = Molarity x Volume

Mole of Na₂SO₄ = 0.3 × 0.25

Mole of Na₂SO₄ = 0.075 mole

<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>
  • Molar mass of Na₂SO₄ = 142.05 g/mol
  • Mole of Na₂SO₄ = 0.075 mole
  • Mass of Na₂SO₄ =?

Mass = mole × molar mass

Mass of Na₂SO₄ = 0.075 × 142.05

Mass of Na₂SO₄ = 10.65 g

Thus, 10.65 g of Na₂SO₄ is needed to prepare the solution.

Learn more about molarity:

brainly.com/question/15370276

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What is the volume of 4.00 moles of CO2 gas at STP?
likoan [24]

Answer:

89.6L

Explanation:

1mole of any gas occupies 22.4L. This simply means that,

1mole of CO2 occupies 22.4L at stp.

Therefore, 4moles of CO2 will occupy = 4 x 22.4 = 89.6L

8 0
3 years ago
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e
notsponge [240]

<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.136-(-0.76)=0.624V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

where,

E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

[Zn^{2+}]=2.5\times 10^{-3}M

[Sn^{2+}] = ?

Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

Hence, the concentration of Sn^{2+} ions is 9.0\times 10^{-3}M

3 0
3 years ago
N2 reacts with hydrogen gas according to the following equation:
slega [8]

Answer:

Mass = 51 g

Explanation:

Given data:

Mass of nitrogen = 41.93 g

Mass of ammonia formed = ?

Solution:

Chemical equation:

N₂ + 3H₂      →       2NH₃

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 41.93 g/ 28 g/mol

Number of moles = 1.5 mol

now we will compare the moles of nitrogen and ammonia.

                N₂          :           NH₃

                  1          :           2

                1.5         :         2/1×1.5 = 3 mol

Mass of ammonia formed:

Mass = number of moles × molar mass

Mass = 3 mol × 17 g/mol

Mass = 51 g

6 0
3 years ago
1.The movement of particles from a less crowded area to a more crowded area requires
Harrizon [31]
1. Energy as particles can't move without kinetic (movement) energy 
2. I think it's Diffusion and Active Transport 
7 0
3 years ago
Determine the volume of atmospheric air (at 14 lb/in.^2) needed to fill your bike tires (assuming it holds 500 mL of air) to the
Fiesta28 [93]

Boyle Law says “the pressure of fixed amount of ideal gas which is at constant temperature is inversely proportional to its volume".<span>

P = 1/V

<span>Where, P is pressure of the ideal gas and V is volume of the ideal gas.</span>

<span>For two situations, this law can be added as;
P</span>₁V₁ = P₂V₂<span>

</span><span>14 lb/in² x V₁ = 70 lb/in² x 500 mL</span><span>
                 </span><span>V₁ = 2500 mL</span><span>

Hence, the needed volume of atmospheric air = 2500 mL

<span>Here, we made two </span>assumptions. They are,
1. The atmospheric air acts as ideal gas.
2. Temperature is a constant.

<span>We didn't convert the units to SI units since converting volume and pressure are products of two numbers, they will cut off. </span></span></span>

3 0
4 years ago
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