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3 years ago

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as

Chemistry

1 answer:

zubka84 [21]3 years ago

6 0

Answer:

Explanation:

MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂

87 g 22.4 x 10³ mL

volume of given chlorine gas at NTP or at 760 Torr and 273 K

= 175 x ( 273 + 25 ) x 715 / (273 x 760 )

= 179.71 mL

22.4 x 10³ mL of chlorine requires 87 g of MnO₂

179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g

= 696.77 x 10⁻³ g

= 696.77 mg .

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Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

$R_{in}$ $= \frac{0.04}{100}*2000$

$R_{in}$ = 0.8

The rate-out

$R_{out}$ = $\frac{A}{6000}*2000$

$R_{out}$ = $\frac{A}{3}$

We can say that:

$\frac{dA}{dt}=$ $0.8-\frac{A}{3}$

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

$\frac{dA}{dt} +\frac{A}{3} =0.8$

Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

so we have:

$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

$\frac{d}{dt}[e^{\frac{1}{3}t}A]$ $= 0.8e^{\frac{1}{3}t$

$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

$C= 12-2.4$

$C =9.6$

Hence;

$A(t) = 2.4 +9.6e^{-\frac{t}{3}}$

$A(0) = 2.4 +9.6e^{-\frac{10}{3}}$

$A(t) = 2.74$

∴ the concentration at 10 minutes is ;

= $\frac{2.74}{6000}*100$ %

= 0.0456667 %

= 0.046% to three decimal places

7 0

4 years ago

What is the mass in grams of 3x10 atoms of helium

Zepler [3.9K]

Answer:

here

Explanation:

What is the total mass of $$3.01 x 10^23$$ atoms of helium gas?

✓ Well $$"Avogadro's number"$$ of helium atoms has a mass of $$4.0*g$$. Explanation: And $$"Avogadro's number"$$

3 0

3 years ago

Enter the net ionic equation for this reaction. Express your answer as a net ionic equation. Identify all of the phases in your

Lisa [10]

Answer:

2H+(aq) + 2OH-(aq) → 2H2O(l)

Explanation:

Step 1: The balanced equation

2HCl(aq)+Ca(OH)2(aq) → 2H2O(l)+CaCl2(aq)

This equation is balanced, we do not have the change any coefficients.

Step 2: The netionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.

2H+(aq) + 2Cl-(aq) + Ca^2+(aq) + 2OH-(aq) → 2H2O(l) + Ca^2+(aq) + 2Cl-(aq)

After canceling those spectator ions in both side, look like this:

2H+(aq) + 2OH-(aq) → 2H2O(l)

6 0

3 years ago

What does RNA do?

MariettaO [177]

Answer:

b, its a polymer that translates genetic information.

7 0

3 years ago

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