What is true about energy

The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV

allsm [11]

Answer:

The rate of change of the temperature is 0.0365 Kelvin per minute.

Explanation:

Step 1: Given data

ideal gas law: P*V = n*R*T

with P= pressure of the gas ( in atm) = 9.0 atm

with V= volume of the gass (in L) =12L

with n = number of moles = 10 moles

R = gas constant = 0.0821 L*atm* K^−1*mo^−1

T = temperature = TO BE DETERMINED

The volume decreases with a rate of 0.17L/min = dV/dT = -0.17

The pressure increases at a rate of 0.13atm/min = dP/dT

Step 2: The ideal gas law

P * [dV/dT] + V * [dP/dT] = nR * dT/dt

9 atm * (-0.17L/min) + 12L * 0.13atm/min = 10 moles * 0.0821 L*atm* K^−1*mo^−1 *dT/dt

0.03 = 0.821 * dT/dt

dT/dt = 0.03/0.821

dT/dt = 0.0365

Since the gas constant is expressed in Kelvin and not in °C, this means that the rate of chagnge of the temperature is 0.0365 Kelvin per 1 minute.

7 0

3 years ago

An irregular chunk of iron has a density of 7.87 g/cm3. A 14.5 gram chunk is added to a graduated cylinder. If the final volume

bulgar [2K]

Answer:

37.96 mL

Explanation:

From the question given above, the following data were obtained:

Density (D) of iron = 7.87 g/cm³

Mass (m) of iron = 14.5 g

Volume of Cylinder + Iron = 39.8 mL

Volume of Cylinder =?

Next, we shall determine the volume of the chunk of iron. This can be obtained as follow:

Density (D) of iron = 7.87 g/cm³

Mass (m) of iron = 14.5 g

Volume (V) of iron =?

D = m/V

7.87 = 14.5 /V

Cross multiply

7.87 × V = 14.5

Divide both side by 7.87

V = 14.5 / 7.87

V = 1.84 cm³

Recall:

1 cm³ = 1 mL

Therefore,

1.84 cm³ = 1.84 mL

Therefore, the volume of the chunk of iron is 1.84 mL.

Finally, we shall determine the starting volume of the cylinder as follow:

Volume of Cylinder + Iron = 39.8 mL

Volume of Iron = 1.84 mL.

Volume of Cylinder =?

Volume of Cylinder = (Volume of Cylinder + Iron) – Volume of Iron

Volume of Cylinder = 39.8 – 1.84

Volume of Cylinder = 37.96 mL

Therefore the starting volume of cylinder is 37.96 mL

3 0

3 years ago

Calculate the hight of a column of liquid glycerol in meters required to exert the same pressure as 3.02 m if CCl4?

katovenus [111]

Answer:

Approximately $3.81\; \rm m$ .

Explanation:

Look up the density $\rho$ of carbon tetrachloride, $\rm CCl_4$ , and glycerol:

Density of carbon tetrachloride: approximately $1.59\times 10^{3}\; \rm kg \cdot m^{-3}$ .
Density of glycerol: approximately $1.26\times 10^{3}\; \rm kg \cdot m^{-3}$ .

Let $g$ denote the gravitational field strength. (Typically $g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth.) For a column of liquid with a height of $h$ , if the density of the liquid is $\rho$ , the pressure at the bottom of the column would be:

$P = \rho\cdot g \cdot h$ .

The pressure at the bottom of this carbon tetrachloride column would be:

$\begin{aligned} P &= \rho \cdot g \cdot h \\ & \approx 1.59\times 10^{3} \; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1} \times 3.02 \; \rm m \approx 4.71 \times 10^{4} \; \rm N \cdot m^{-2} \end{aligned}$ .

Rearrange the equation $P = \rho\cdot g \cdot h$ for $h$ :

$\displaystyle h = \frac{P}{\rho \cdot g}$ .

Apply this equation to calculate the height of the liquid glycerol column:

$\begin{aligned}h &= \frac{P}{\rho \cdot g} \\ &\approx \frac{4.71 \times 10^{4}\; \rm N \cdot m^{-2}}{1.26 \times 10^{3}\; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1}} \approx 3.81\; \rm m\end{aligned}$ .

4 0

3 years ago

Which is a compound that contains carbon?

Anastaziya [24]

carbon dioxide, that is what I found

7 0

3 years ago

Which line represents a stationary object.

tiny-mole [99]

The green one cause its staying in the same position even when time is passing. - hope this helped have a great day

7 0

3 years ago

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