Asealed-tube manometer can be used to measure pressures below atmospheric pressure. the tube above the mercury is evacuated. when there is a vacuum in the flask, the mercury levels in both arms of the u-tube are equal. if a gaseous sample is introduced into the flask, the mercury levels are different. the difference 'h' is a measure of the pressure of the gas inside the flask. if 'h' is equal to 6.5cm, calculate the pressure in the flask in torr, pascals, and atmospheres. (i need trying to figure out how to get the answer, not what the answer is.)

dependent variable the right one is d

Explanation:

i just did it

The Answer is d

Explanation:

The correct answers are:

6.5 cm = 65.00001052 torr

6.5 cm = 8665.93 Pa

6.5 cm = 0.08552632119 atm

Explanation

The conversion rate from centimeters mercury (cm Hg) to torr is

1 torr = 0.099999983814461 cm.

We have 6.5 cm; this means to convert we divide:

6.5/0.099999983814461 = 65.00001052

The conversion rate from cm Hg to Pascals (Pa) is

1 cm = 1333.22 Pa

We have 6.5 cm; this means to convert we multiply:

6.5(1333.22) = 8665.93

The conversion rate from cm Hg to atmospheres (atm) is

1 atm = 75.999995199606 cm

We have 6.5 cm; this means to convert we divide:

6.5/75.999995199606 = 0.08552632119

3. 9.84 L.

4. 0.1211 mol.

5. Barometer, Bourdon tube, and Aneroid barometers.

6. 4.0 K.

7. 310 K.

8. 1.0 L.

9. 2.637 L.

Explanation:

3. Calculate the approximate volume of a 1.50 mol sample of gas at 300K and a pressure of 3.75 atm.

To calculate the no. of moles of a gas, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 3.75 atm).

V is the volume of the gas in L.

n is the no. of moles of the gas in mol (n = 1.5 mol).

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K (T = 300 K).

∴ V = nRT/P = (1.5 mol)( 0.082 L.atm/mol.K)( 300 K)/(3.75 atm) = 9.84 L.

4. Pressure CO2 = 760 mm Hg Volume CO2 = 2.965 L Temperature CO2= 25.5 °C . How many moles of CO2 were produced?

∵ P = 760 mmHg = 1.0 atm, V = 2.965 L, T = 25.5°C + 273 = 298.5 K, R = 0.082 L.atm/mol.K.

∴ n = PV/RT = (1.0 atm)(2.965 L)/(0.082 L.atm/mol.K)(298.5 K) = 0.1211 mol.

5. What instrument do you measure pressure with?

Barometer, Bourdon tube, and Aneroid barometers.

A barometer is a scientific instrument used to measure air pressure. Pressure tendency can forecast short term changes in the weather. Many measurements of air pressure are used within surface weather analysis to help find surface troughs, high pressure systems and frontal boundaries. Another type is the Bourdon tube. It consists of a flattened tube that is shaped like a “C.” When the pressure outside the tube is greater than the pressure inside, the “C” curls up more. When the pressure inside is greater the “C” uncurls more. A third type is Aneroid barometers. They use vacuum capsules called cells. As air pressure changes, the cell expands and contracts, and a device measure this.

6. The volume of a gas is 500 mL at 20.0 K. What will be the new temperature if the gas is compressed to 100 mL?

If n and P are constant, and have two different values of V and T:

V₁T₂ = V₂T₁

V₁ = 500 mL, T₁ = 20 K.

V₂ = 100 mL, T₂ =??? K.

∴ T₂ = V₂T₁/V₁ = (100 mL)(20 K)/(500 mL) = 4.0 K.

7. The volume of a sample of oxygen is 200.0 mL when the pressure is 4.0 atm and the temperature is 37.0 °C. What is the new temperature if the volume increases to 400.0 mL and the pressure decreases to 2.000 atm?

If n is constant, and have two different values of (P, V and T):

P₁V₁T₂ = P₂V₂T₁

P₁ = 4.0 atm, V₁ = 200 mL, T₁ = 37°C + 273 = 310 K.

P₂ = 2.0 atm, V₂ = 400 mL, T₂ =??? K.

∴ T₂ = P₂V₂T₁/P₁V₁ = (2.0 atm)(400 mL)(310 K)/(4.0 atm)(200 mL) = 310 K.

8. The pressure on a 2.00L sample of CO2(g) is increased from 50 kPa to 100 kPa. What is the new volume of the gas?

If n and T are constant, and have two different values of V and P:

P₁V₁ = P₂V₂

P1 = 50 kPa, V1 = 2.0 L.

P₂ = 100 kPa, V₂ = ??? L.

∴ V₂ = P₁V₁/P₂ = (50 kPa)(2.0 L)/(100 kPa) = 1.0 L.

9. The temperature of a 2.0 liter sample of helium gas at STP is increased to 37°C, and the pressure is decreased to 80 kPa. What is the new volume of the helium sample?

If n is constant, and have two different values of (P, V and T):

P₁V₁T₂ = P₂V₂T₁

P₁ = 1.0 atm (standard P), V₁ = 2.0 L, T₁ = 25°C + 273 = 298 K (standard T).

P₂ = 80 kPa = 0.789 atm, V₂ = ??? L, T₂ = 37°C + 273 = 310 K.

V₂ = P₁V₁T₂/P₂T₁ = (1.0 atm)(2.0 L)(310 K)/(0.789 atm)(298 K) = 2.637 L.

the one that starts with d and the one that starts with q

Explanation:

This problem employs conversion of units.

1cm Hg=10 mmHg

1mmHg= 1 Torr

1 atm = 760mmHg

1 atm = 101,325 Pa

6.5cmHg*(10mmHg/1cmHg)= 65 mmHg = 65 Torr

65mmHg*(1 atm/760mmHg)= 0.086 atm

0.086atm*(101,325 Pa/1 atm)= 8665.95 Pa

The variable "air pressure inside the football" in this experiment is quantitative and dependent variable.

Explanation:

In the experiment, the scientist filled the foot balls with fix quantity of pressure which then kept for 12 hours . After 12 hours the pressure inside the footballs was again measured. Pressure here is a quantitative variable.

Since, air of temperature is an independent variable. Pressure being dependent variable depends upon the temperature of the air.

Independent variable ; The variable which does not depends upon any other variable.

Dependent variable ; The variable which depends upon independent variable or varies with change independent variable..

Qualitative variables are are the variables which cannot be expressed in form of numbers. They generally convey the characteristic ,category ,quality or type.

Where as Quantitative variable are the variables which can be expressed in the form of number.These are the variables which exist along the continuous sequence.