Anormal breath of air is about 1.00 l. if the air is drawn in at 18.0 ∘c and is warmed to 35.0 ∘c, what will be the new volume of the air? assume that the pressure and amount of the gas remain constant.

The new volume of the air is 1.05 liters.

Explanation:

By Charles's law, we can make a relation between temperature and volume, where the volume of a gas (always in constant pressure) is directly proportional to its absolute temperature (T° in K)

So, when the temperature is increasing, volume increases

and when the temperature decreases, volume decreases.

V1/T1 = V2/T2

T° in K = T ° in C + 273

18°C + 273 = 291 K

35°C + 273 =308 K

1 L/291K = V2/308K

(1L /291K). 308K = V2

1.05 L = V2

output?

explanation:

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.