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solmaris [256]
2 years ago
14

If a gas has a molecular mass of 44.0, the volume of 88.0 grams of the gas at STP would be ..

Chemistry
1 answer:
Montano1993 [528]2 years ago
3 0
2) 11.2
All you have to do is multiply 44 by 22.4 L, which equals 985.6
Then you divide 985.6 by 88 to get your answer of 11.2 L
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NaOH has a molarity of 2.5 M and a volume of 4 Liters, how many moles are in<br> the solution?
Inga [223]

Answer:

10. mol NaOH

Explanation:

2.5 M = 2.5 mol/L

2.5 mol/L * 4 L = 10. mol

7 0
2 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
A potted plant is placed under a grow lamp, which provides 6,400 J of energy to the plant and the soil over the course of an hou
STatiana [176]

Answer:

823.7g

Explanation:

Using the formula as follows:

Q = m × c × ∆T

Where;

Q = amount of heat (J)

m = mass of substance (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

Using the information given in this question as follows:

Q = 6,400 J

m = ?

c of soil = 0.840 J/g°C

∆T = 9.25°C

Using Q = mc∆T

m = Q ÷ c∆T

m = 6,400 ÷ (0.840 × 9.25)

m = 6400 ÷ 7.77

m = 823.7g

8 0
3 years ago
Read 2 more answers
Which pure substance is made of only one kind of atom?
aleksley [76]

Answer:

the answer is D i believe. i am not shure but im like 70% sure

7 0
3 years ago
How many liters of o2 do you have if you have 5.8 moles of o2
polet [3.4K]

Answer:

130 Liters

Explanation:

if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)

3 0
3 years ago
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