All categories

3 years ago

I need help with part B . please

Chemistry

1 answer:

myrzilka [38]3 years ago

8 0

So it's good to map out what you know you have and work from there:
We have two liter measurements and one mole measurement, and we need to find the moles.

For this problem, think of it this way: 46 liters of gas = 1.4 moles.
If one side changes, the other has to as well (if the liters decrease, the moles decrease. if the liters increase, so do the moles.) What you can do is put this into a fraction:

 1.4 moles
 46 L 

if we know that each liter of gas is equal to x amount of moles, we know that 11.5 liters equals some amount of moles. You can put this into a fraction too, and make it equal to the other fraction:

1.4 moles = x moles
46 L 11.5 L

Then get your calculator out and do some algebra.

11.5 * (1.4/46) = x

The answer should come out to be: 0.35 moles

You might be interested in

How do you solve questions 6 and 7

bearhunter [10]

I know the answer to your question but if you want me to answer this question first, I need your help with my question "How do I solve number 12? Quick please I need help ASAP!!!"

6 0

3 years ago

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

3 years ago

Law of Conservation of Mass in terms of atoms

VARVARA [1.3K]

No atoms are lost or made during the chemical reaction so the total mass of the products is equal to the total mass of the reactants. In an atom, protons and neutrons contribute to the mass and since the number of them doesn’t change, the mass doesn’t either.

4 0

3 years ago

Copper is malleable, ductile, and conducts heat and electricity. Would you classify copper as a metal, nonmetal, or metalloid? W

Morgarella [4.7K]

Answer:

The elements can be classified as metals, nonmetals, or metalloids. Metals are good conductors of heat and electricity, and are malleable (they can be ... and electricity, and are not malleable or ductile; many of the elemental nonmetals are ... under certain circumstances, several of them can be made to conduct electricity.

Hope this helps!

Explanation:

5 0

3 years ago

Read 2 more answers

Solid aluminum and gaseous oxygen read in a combination reaction to produce aluminum oxide. 4Al(s) + 3O_2(g) rightarrow 2Al_2O_3

Kryger [21]

Answer:

4.7 g. Option 5 is the right one.

Explanation:

4Al(s) + 3O₂ (g) ⇄ 2Al₂O₃ (s)

We convert the mass of reactants to moles, in order to determine the limiting.

2.5 g Al / 26.98 g/mol → 0.092 moles of Al

2.5 g O₂ / 32g/mol → 0.078 moles of O₂

Ratio is 4:3. 4 moles of Al react with 3 moles of O₂

Then, 0.092 moles of O₂ would react with (0.092 . 3)/ 4 = 0.069 moles O₂

We have 0.078 moles of O₂ and we need 0.069 moles, the oxygen is the limiting in excess. Therefore the Al is the limiting reactant.

Ratio is 4:2. 4 moles of Al, can produce 2 moles of Al₂O₃

Then, 0.092 moles of Al would produce (0.092 .2) / 4 = 0.046 moles

If we convert the moles to mass, we find the anwer:

0.046 mol . 101.96 g/mol = 4.69 g

5 0

3 years ago