Ask question

Ask question

All categories

3 years ago

7

If the partial pressures of the three gases in the tank are 38.39 atm of O2, 3.38 atm of He, and 25.33 atm of N2, what is the to

tal partial pressure in the tank

1 answer:

Tasya [4]3 years ago

4 0

Answer:

67.1 atm

Explanation:

From Dalton's law of Partial Pressures, the total partial pressure equals the sum of the individual partial pressures.

P = P₁ + P₂ + P₃

P₁ = 38.39 atm of O₂

P₂ = 3.38 atm of He

P₃ = 25.33 atm of N₂

So, P = P₁ + P₂ + P₃ = 38.39 atm + 3.38 atm + 25.33 atm = 67.1 atm

You might be interested in

Why do astronomers use the word on to describe angles on the sky rather than angles in the sky?

Sever21 [200]

<span>The use of the word on instead of the word in when referring to the angular distance between celestial objects comes about because all of the objects appear to be on the celestial sphere and at an indeterminable distance. While we know that objects are at different distances in the sky, their distance from Earth is irrelevant in determining the angular distance between the two objects as viewed from Earth.</span>

8 0

2 years ago

A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in

Kruka [31]

Answer:

1020g

Explanation:

Volume of can= $1100cm^3=1100\times 10^{-6}m^3$

$1cm^3=10^{-6}m^3$

Mass of can=80g= $\frac{80}{1000}=0.08kg$

1Kg=1000g

Density of lead= $11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3$

By using $1g/cm^3=10^3kg/m^3$

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can and lead inside it they are floating in the water.

Buoyancy force = $F_b=Weight of can+weight of lead$

$\rho_wV_cg=m_cg+m_lg$

Where $\rho_w=10^3kg/m^3=$ Density of water

$m_c=$ Mass of can

$m_l=$ Mass of lead

$V_c=$ Volume of can

Substitute the values then we get

$1000\times 1100\times 10^{-6}=0.08+m_l$

$1.1-0.08=m_l$

$m_l=1.02 kg=1.02\times 1000=1020g$

$1 kg=1000g$

Hence, 1020 grams of lead shot can it carry without sinking water.

4 0

3 years ago

Which one is it? Help ASAP

dezoksy [38]

Answer:

extreme heat, because no physical damage can demagnetize a magnet

Explanation:

6 0

2 years ago

Read 2 more answers

At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v

Cerrena [4.2K]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.

$R= 6370*10^3 m$

$v = 239m/s$

$a = 16.5m/s^2$

The circumference of the earth would be

$\phi = 2\pi R$

Velocity is defined as,

$v = \frac{x}{t}$

$t = \frac{x}{v}$

Here $x = \phi$ , then

$t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{239}$

$t = 167463.97s$

Therefore will take 167463.97 s or 1 day 22 hours 31 minutes and 3.97seconds

4 0

2 years ago

Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length

SVEN [57.7K]

Answer with Explanation:

We are given that

Length of wire 1= $L_1$

Length of wire 2= $L_2$

Resistivity of copper wire= $\rho_1=1.7\times 10^{-5}\Omega-m$

Resistivity of aluminum wire= $\rho_2=2.82\times 10^{-5}\Omega-m$

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

Resistance = $\frac{\rho l}{A}$

When diameter of wires are same then their cross section area are also same .

$l=\frac{RA}{\rho}$

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity is small then the length of wire will be large.

$\rho_1$

Therefore, $L_1>L_2$

Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

$\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66$

$L_1=1.66L_2$

7 0

2 years ago