Which of the following equations are balanced?

An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h

vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base = w

Height of the pyramid = h

Volume of the pyramid = $V=\frac{1}{3}lwh$

We have:

Rate at which water is filled in cube = $\frac{dV}{dt}= 45 cm^3/s$

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

$V=\frac{1}{3}\times l^2\times h$

$\frac{l}{h}=\frac{6}{13}$

$l=\frac{6h}{13}$

$V=\frac{1}{3}\times (\frac{6h}{13})^2\times h$

$V=\frac{12}{169}h^3$

Differentiating V with respect to dt:

$\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}$

$\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}$

Putting, h = 4 cm

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}$

$=13.20 cm/s$

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

3 0

3 years ago

32.7 grams of water vapor takes up how many liters at standard temperature and pressure (273 K and 100 kPa)?

kotegsom [21]

Under standard temperature and pressure conditions, it is known that 1 mole of a gas occupies 22.4 liters.

From the periodic table:
molar mass of oxygen = 16 gm
molar mass of hydrogen = 1 gm
Thus, the molar mass of water vapor = 2(1) + 16 = 18 gm

18 gm of water occupies 22.4 liters, therefore:
volume occupied by 32.7 gm = (32.7 x 22.4) / 18 = 40.6933 liters

5 0

3 years ago

Potassium decays by beta decay part of the nuclear equation is shown below. fill in the blank with a number

andre [41]

Answer:

20

Explanation:

5 0

3 years ago

Read 2 more answers

Which action changes the identity of substance referenced

Scilla [17]

Answer:

When iron turns to rust, the chemical identity of the iron changes. When you break ice, the ice is still ice

5 0

3 years ago

A particular reactant decomposes with a half‑life of 113 s when its initial concentration is 0.331 M. The same reactant decompos

algol13

Answer:

The reaction is second-order, and k = 0.0267 L mol^-1 s^-1

Explanation:

Step 1: Data given

The initial concentration is 0.331 M

half‑life time = 113 s

The same reactant decomposes with a half‑life of 243 s when its initial concentration is 0.154 M.

Step 2: Determine the order

The reaction is not first-order because the half-life of a first-order reaction is independent of the initial concentration:

t½ = (ln(2))/k

Calculate k for the two conditions given:

⇒ 113 s with initial concentration is 0.331 M

t½ = ([A]0)/2k

113 s = (0.331 M)/2k

k = 0.00146 mol L^-1 s^-1

⇒ 243 s with an initial concentration is 0.154 M

t½ = ([A]0)/2k

243 s = (0.154 M)/2k

k = 0.000317 mol L^-1 s^-1

The values of k are different, so that rules out zero-order.

Step 3: Calculate if it's a second-order reaction

For a second-order reaction, the half-life is given by the expression

t½ = 1/((k*)[A]0))

Calculate k for the two conditions given:

⇒ 113 s when its initial concentration is 0.331 M

t½ = 1/((k*)[A]0))

113 s = 1/(k*(0.331 M))

k = 1/((0.331 M)*(113 s)) = 0.0267 L mol^-1 s^-1

⇒ 243 s when its initial concentration is 0.154 M

t½ = 1/((k*)[A]0))

243 s = 1/(k*(0.154 M))

k = 1/((0.154 M)*(243 s)) = 0.0267 L mol^-1 s^-1

The values of k are the same, so the reaction is second-order, and k = 0.0267 L mol^-1 s^-1

4 0

4 years ago