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Sphinxa [80]
3 years ago
11

How does the mass of a gas particle affect its rate of effusion?

Chemistry
1 answer:
ryzh [129]3 years ago
4 0
The mass of gas particle affects the rate of effusion by decreasing the rate as the mass increases. This describe by the Graham's Law of Effusion where the rate of effusion is inversely proportional to the square root of the molar mass of a substance. Hope this helps.
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If an object is accelerating the forces acting on the object are ?​
belka [17]

Answer:

If an object is accelerating the forces acting on the object are BALANCED.

Explanation

if an object is moving at a constant rate of acceleration, the the forces acting upon it are balanced .

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3 years ago
How does air flow when a low-pressure center occurs in the<br> atmosphere?
Monica [59]

Answer:

A

Explanation:

It is so because the low air pressure create vacuum and the air from high pressure area move toward the low air pressure.

8 0
2 years ago
Calculate the [OH-]<br> [H+]=0.000025
Harman [31]

Explanation:

There are several ways to define acids and bases, but pH and pOH refer to hydrogen ion concentration and hydroxide ion concentration, respectively. The "p" in pH and pOH stands for "negative logarithm of" and is used to make it easier to work with extremely large or small values. pH and pOH are only meaningful when applied to aqueous (water-based) solutions. When water dissociates it yields a hydrogen ion and a hydroxide.

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3 years ago
(10 points) easy WILL GIVE BRAINLIEST
nignag [31]

Answer:

Both b/c a chemical formula tells you how many and a sketch formula shows how they are bonded together.

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4 0
2 years ago
Read 2 more answers
A 50.0 g sample of scandium, sc, is heated by exposure to 1.50 x 10 3 j. The temperature of the sc is raised by 61.1 o
statuscvo [17]

Given mass of Scandium = 50.0 g

Increase in temperature of the metal when heated = 61.1^{0}C

Heat absorbed by Scandium = 1.50*10^{3}J

The equation showing the relationship between heat, mass, specific heat and temperature change:

Q = m C (deltaT)

Where Q is heat = 1.50*10^{3}J

m is mass = 50.0 g

ΔT = 61.1^{0}C

On plugging in the values and solving for C(specific heat) we get,

1.50*10^{3}J=50.0g(C)(61.1^{0}C)

C = 0.491\frac{J}{g^{0}C }

Specific heat of the metal = 0.491\frac{J}{g^{0}C }

7 0
3 years ago
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