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Talja [164]
3 years ago
11

How many moles of oxygen gas (O2) are in a 2.50 L container at standard temperature and pressure?

Chemistry
1 answer:
Anon25 [30]3 years ago
6 0

Hey there!:

1 mole O2 ----------------- 22.4 L  ( at STP )

( moles O2 ) -------------- 2.50 L

moles O2  = 2.50 * 1 / 22.4

moles O2 = 2.50 / 22.4

moles O2 =  1.12*10⁻¹


Answer B


Hope that helps!

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The pressure in car tires is often measured in pounds per square inch ( lb/in.2 ), with the recommended pressure being in the ra
Goshia [24]

when we convert 32.5 lb/in² to atmosphere, the result obtained is 2.21 atm

<h3>Conversion scale</h3>

14.6959 lb/in² = 1 atm

<h3>Data obtained from the question</h3>
  • Pressure (in lb/in²) = 32.5 lb/in²
  • Pressure (in ATM) =?

<h3>How to convert 32.5 lb/in² to atm</h3>

14.6959 lb/in² = 1 atm

Therefore

32.5 lb/in² = 32.5 / 14.6959

32.5 lb/in² = 2.21 atm

Thus, 32.5 lb/in² is equivalent to 2.21 atm

Learn more about conversion:

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4 0
1 year ago
How do the lithosphere and the asthenosphere work together?
miskamm [114]
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5 0
2 years ago
The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol
Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration PCl_3 and Cl_2.

\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}

\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M

\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}

\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M

The given equilibrium reaction is,

                            PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initially                 0.70        0.70              0

At equilibrium    (0.70-x)   (0.70-x)           x

The expression of K_c will be,

K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}

K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}

Now put all the given values in the above expression, we get:

49=\frac{(x)}{(0.70-x)\times (0.70-x)}

By solving the term x, we get

x=0.59\text{ and }0.83

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of PCl_3 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of Cl_2 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of PCl_5 at equilibrium = x = 0.59 M

Therefore, the concentration of PCl_3 at equilibrium is 0.11 M

3 0
3 years ago
2567 milliliters to liters
andrew-mc [135]

Answer:

2.567 litres

Explanation:

I dont have a proper explanation sorry

6 0
2 years ago
The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is
Art [367]
Hi! Well the formula of kinetic energy is Ke = 0.5 x M x V^2
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