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Effectus [21]
1 year ago
15

If 2.0 mol of gas a is mixed with 1.0 mol of gas b to give a total pressure of 1.6 atm, what is the partial pressure of gas a an

d b?
Physics
2 answers:
devlian [24]1 year ago
8 0
<span>PA = 1.06 atm and PB = 0.53 atm</span>
jarptica [38.1K]1 year ago
7 0

Answer:

Gas A=1,06 atm

Gas B=,53 atm

Explanation:

YOu have 3 moles of two different gases mixed into an specific container, the partial pressure is the pressure that each one of them would have if contained in the same container by themselves, as you have 3 moles of substance you just divde 1,6 atm by the 3 moles and assign each gas its correspondent pressure:

1,6/3=,53

Gas A=1,06 atm

Gas B=,53 atm

You might be interested in
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore  P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

7 0
2 years ago
When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical powe
ch4aika [34]

Answer:

4.41 W

Explanation:

P = IV, V = IR

P = V² / R

Given that P = 0.0625 when V = 1.50:

0.0625 = (1.50)² / R

R = 36

So the resistor is 36Ω.

When the voltage is 12.6, the power consumption is:

P = (12.6)² / 36

P = 4.41

So the power consumption is 4.41 W.

5 0
2 years ago
A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber
Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}

Now calculate the bouyant force acting on the lead

F_L = Vpg

F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N

This force will act in upward direction

Gravitational force on the lead due to its mass  will act in downward direction

Hence the difference of this two force

T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N

If V is the volume submerged in the water then bouyant force on the bobber is

F_B=V'pg

Equate bouyant force with the tension and gravitational force

F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}

Now Total volume of bobble is

\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\

=\large\boxed{4.52 \%}

7 0
2 years ago
A car travels straight for 20 miles on a road that is 30° north of east. What is the east component of the car’s displacement to
Virty [35]
17.3 would be the correct answer.

4 0
2 years ago
Read 2 more answers
A split highway has a number of lanes for traffic. For traffic going in one direction, the radius for the inside of the curve is
Arte-miy333 [17]
Cuál es tu pregunta
7 0
1 year ago
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