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3 years ago

A person breathes 2.6 L of air at -11 C into her lungs, where it is warmed to 37 C. What is its new Volume?

Chemistry

1 answer:

GaryK [48]3 years ago

7 0

<h3><u>Answer</u>;</h3>

= 3.076 L

<h3><u>Explanation;</u></h3>

Using the Charles' Law;

V1/T1 =V2/T2

V1 =2.6 L, T1 = (-11 + 273) K =262 K

V2 = ? , T2 = (37 +273) k = 310 K

Therefore;

2.6/262 = V2/310

V2 = (2.6 × 310)/262

<u> = 3.076 L </u>

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Consider the reaction below: 3 h2(g) + n2(g) ? 2 nh3(g) if there are 6 mol of nitrogen (n2) and more than enough hydrogen (h2),

Alex73 [517]

1 mol N₂ - 2 mol NH₃
6 mol N₂ - x mol NH₃

x=2×6/1=12 mol

12 mol NH₃

3 0

3 years ago

The metric unit for volume is the? <br><br> A. pound<br> B.quart<br> C.liter<br> D.gallon

Vladimir [108]

I believe it is liter.

5 0

4 years ago

Read 2 more answers

mixture or pure substance: 1.blood 2.dyes 3.self-raising flour 4.muesli 5.copper wire 6.distilled water 7.table salt 8.milk 9.br

jenyasd209 [6]

Answer: Mixture: Blood , Self raising flour,muesli ,dyes, milk, tea, air, bronze

Pure substance: Copper wire, distilled water, table salt, oxygen.

Explanation:

Mixture is a substance which is made up two or more number of compounds which chemically inactive and retain their distinct chemical properties.

Blood , Self raising flour,muesli ,dyes, milk, tea, air, bronze

Pure substance is defined as anything with uniform and unchanging composition is known s pure substance.

Copper wire, distilled water, table salt, oxygen.

5 0

3 years ago

To 100.0 g water at 25.00 ºc in a well-insulated container is added a block of aluminum initially at 100.0 ºc. the temperature o

11111nata11111 [884]

When the amount of heat gained = the amount of heat loss

so, M*C*ΔTloses = M*C* ΔT gained

when here the water is gained heat as the Ti = 25°C and Tf= 28°C so it gains more heat.

∴( M * C * ΔT )W = (M*C*ΔT) Al

when Mw is the mass of water = 100 g

and C the specific heat capacity of water = 4.18

and ΔT the change in temperature for water= 28-25 = 3 ° C

and ΔT the change in temperature for Al = 100-28= 72°C

and M Al is the mass of Al block

C is the specific heat capacity of the block = 0.9

so by substitution:

100 g * 4.18*3 = M Al * 0.9*72

∴ the mass of Al block is = 100 g *4.18 / 0.9*72

= 19.35 g

4 0

3 years ago

A sample of xenon gas occupies a volume of 6.80 L at 52.0°C and 1.05 atm. If it is desired to increase the volume of the gas sam

Pavlova-9 [17]

Answer:

207.03°C

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

°C = 207.03°C

Therefore, the final temperature of the gas will be 207.03°C

5 0

3 years ago