A gas has a pressure of 3.33 kPa at 343 K. What will the new pressure be at 655 K if the volume does not change?

Add them to your friend's list and have them support you along your Brainly journey. Just go to the profile of the person you’d

Makovka662 [10]

Answer:

yup that's how you make friends with people on brainly

8 0

3 years ago

Which of the following is a unit of length? O A. A liter O B. A kilogram C. A meter O D. A degree

sergeinik [125]

Answer:

C.) A meter

Explanation:

5 0

3 years ago

the carbon atom which becomes asymmetric when the straight chain form of monosaccharide change into ring form is known as____\

amid [387]

Answer:

CARBOHYDRATES AND CARBOHYDRATE METABOLISM

4 0

2 years ago

Calculate the hydroxide ion concentration of a 0.200 M solution of HCIO4

Aleksandr [31]

The concentration of hydroxide ion is 5 $\times$ 10^ − 14 M.

<u>Explanation:</u>

Consider the equilibrium of this acid's dissociation,

H C l O 4 ⇌ H + + C l O 4 -

Moreover, let's assume that H C l O 4 is a strong acid and will fully dissociate.

Hence,

[ H + ] = 0.20 M

Now, recall,

K w = [ H + ] [ O H − ] = 1.0 $\times$ 10 ^− 14

Hence,

⇒ [ O H − ] = K w / [ H + ] = 5 $\times$ 10^ − 14 M.

7 0

4 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