A chemist must dilute 73.9 mL of 400 mM aqueous sodium carbonate solution until the concentration falls to 125 mM . He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits.

Answer:

The final volume of the solution will be 0.236 L.

Explanation:

Concentration of sodium carbonate solution before dilution = $M_1= 400 mM$

Volume of sodium carbonate solution before dilution = $V_1=73.9 mL$

Concentration of sodium carbonate solution after dilution = $M_2= 125 mM$

Volume of sodium carbonate solution after dilution = $V_2=?$

Dilution equation is given by:

$M_1V_1=M_2V_2$

$V_2=\frac{M_1V_1}{M_2}$

$V_2=\frac{400 mM\times 73.9 mL}{125 mM}= 236.48 mL\approx 236 mL$

1 mL = 0.001 L

236 mL = 0.236 L

The final volume of the solution will be 0.236 L.

3 0

3 years ago

How much<br>much hydrogen gas evolved<br>when 1.5 current is passed through water for 1.5 hours?

evablogger [386]

0.042 moles of Hydrogen evolved

<h3>Further explanation</h3>

Given

I = 1.5 A

t = 1.5 hr = 5400 s

Required

Number of Hydrogen evolved

Solution

Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.

Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)

Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻

Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)

So at the cathode H₂ gas is produced

Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C

$\tt mol~e^-=\dfrac{Q}{96500}$

Q = i.t

Q = 1.5 x 5400

Q = 8100 C

mol e⁻ = 8100 : 96500 = 0.084

From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042

4 0

3 years ago

The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effusi

Norma-Jean [14]

Answer:

63.6g/mol

Explanation:

Use the equation of effusion:

47.8/24= $\sqrt{x/16.04}$

solve for x you get 63.6g/mol

3 0

3 years ago

Water (2850 g ) is heated until it just begins to boil. If the water absorbs 5.53×105 j of heat in the process, what was the ini

NARA [144]

Answer is 54 °C.

<em>Explanation;</em>

We can simply use heat equation

Q = mcΔT

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

Let's assume that the initial temperature is T.

Q = 5.53 × 10⁵ J

m = 2850 g

c = 4.186 J/g °C

ΔT = (100 - T) °C <em>Since the water is boiling, the final temperature is 100 °C.</em>

By applying the equation,

5.53 × 10⁵ J = 2850 g x 4.186 J/g °C x (100 - T) °C

(100 - T) °C = 5.53 × 10⁵ J / (2850 g x 4.186 J/g °C )

(100 - T) °C = 46.35 °C

T = 100 - 46.35 C = 53.65 °C

≈ 54 °C

7 0

4 years ago