A 4.44 l container holds 15.4 g of oxygen at 22.55°c. what is the pressure?

1 answer:

5 0

We will use the ideal gas equation:
PV = nRT, where n is moles and equal to mass / Mr
P = mRT/MrV
P = 15.4 x 8.314 x (22.55 + 273) / 32 x 4.44
P = 266.3 kPa

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The fastest pitched baseball was clocked at 46 m/s. assume that the pitcher exerted his force (assumed to be horizontal and cons

Zielflug [23.3K]

Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance

We know that vi = 0 because the ball was at rest initially.

Therefore,

Solving Eq.1 for acceleration,

 v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·s
Substituting the values,
a = 46²/2×1
a = 1058 m/s²

Now applying Newton's 2nd law of motion,

F = ma
= 0.145×1058

F = 153.4 N

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2 years ago

I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......

pav-90 [236]

Answer:

Explanation:

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100 $cm^{2}$ x 10cm = 1000 $cm^{3}$

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

$\frac{1g}{cm^{3} }$ x 1000 $cm^{3}$ = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

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20g ÷ $\frac{8g}{cm^{3} }$ = 2.5 $cm^{3}$

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 $cm^{3}$ overflowed. So now we the same process as in number a) just with a few adjustments.

$\frac{1g}{cm^{3} }$ x (1000 $cm^{3}$ - 2.5 $cm^{3}$ ) = 997.5g

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5 0

3 years ago

A bowl contains 4 red balls and 2 green balls. Two balls are selected simultanesouly and at random. What is the conditional prob

AlladinOne [14]

Answer:0.6

Explanation:

Given

Bowl contains 4 red and 2 Green balls

Probability of selecting two red balls given atleast one of the ball is red + Probability of selecting two red balls given that no ball is red =1

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