All categories

3 years ago

When the pressure that a gas exerts

Chemistry

2 answers:

Mademuasel [1]3 years ago

8 0

Answer:

2.13

Explanation:

Ronch [10]3 years ago

5 0

Answer:

The pressure changes from 2.13 atm to 1.80 atm.

Explanation:

Given data:

Initial pressure = ?

Final pressure = 1.80 atm

Initial temperature = 86.0°C (86.0 + 273 = 359 K)

Final temperature = 30.0°C (30+273 =303 K)

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

P₁ = P₂T₁ /T₂

P₁ = 1.80 atm × 359 K / 303 K

P₁ = 646.2 atm. K /303 K

P₁ = 2.13 atm

The pressure changes from 2.13 atm to 1.80 atm.

You might be interested in

Please help will mark you brainiest! Use word bank when answering the question!

zheka24 [161]

Answer:

Blank 1: varying

Blank 2: constant

Explanation:

7 0

3 years ago

WHAT IT IZ BRAINLY anyways I got a question for y’all do you think someone can forget who they rlly loved ? Can you ever forget

alexdok [17]

Answer:

Its a 50% chance of it happeing

Explanation:

5 0

3 years ago

Anions are formed when *

soldier1979 [14.2K]

Anions are negative ions so it’s when an atom gains electrons

6 0

3 years ago

WHAT BLOCK IN THE PICTURE HAS:

GREYUIT [131]

Answer:

the highest is "A" and the lower is "C"

7 0

3 years ago

Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo

Romashka-Z-Leto [24]

Answer:

$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

Explanation:

Volume of a cone:

$\displaystyle V=\frac{1}{3} \pi r^2 h$

We have $\displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec}$ and we want to find $\displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ?$ when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

$\displaystyle V =\frac{1}{3} \pi (5h)^2 h$
$\displaystyle V =\frac{1}{3} \pi \ 25h^3$

Differentiate this equation with respect to time t.

$\displaystyle \frac{dV}{dt} =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}$
$\displaystyle \frac{dV}{dt} =25 \pi h^2 \ \frac{dh}{dt}$

Plug known values into the equation and solve for dh/dt.

$\displaystyle 10 = 25 \pi (2)^2 \ \frac{dh}{dt}$
$\displaystyle 10 = 100 \pi \ \frac{dh}{dt}$

Divide both sides by 100π to solve for dh/dt.

$\displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}$
$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0

3 years ago