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3 years ago

When might a large volume of material have little mass

1 answer:

4 0

A large volume of material will have a small amount of mass when the material in question is gaseous or porous, and therefore has a lot of space within it despite taking up a large amount of space overall.

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2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Alex_Xolod [135]

Answer: The concentration of hydrogen gas at equilibrium is 0.0275 M

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of HI = 0.550 moles

Volume of container = 2.00 L

$\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M$

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial: 0.275

At eqllm: 0.275-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$K_c=0.0156$

Putting values in above expression, we get:

$0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

4 0

3 years ago

The volume of a sample of oxygen is 200.0 mL when the pressure is 3.000 atm and the temperature is 37.0 oC. What is the new temp

attashe74 [19]

Answer:

Explanation:

Given:

V1 = 200 ml

T1 = 20 °C

= 20 + 273

= 293 K

P1 = 3 atm

P2 = 2 atm

V2 = 400 ml

Using ideal gas equation,

P1 × V1/T1 = P2 × V2/T2

T2 = (2 × 400 × 293)/200 × 3

= 234400/600

= 390.67 K

= 390.67 - 273

= 117.67 °C

7 0

3 years ago

Which is the correct formula for phosphorus pentachloride?

klio [65]

PCl5

hope this helps

have a wonderful day

5 0

3 years ago

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Lunna [17]

Answer:

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3 years ago

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Will give brainlsit need answers asap!

k0ka [10]

Answer:

120 g of NaCl in 300 g H20 at 90 C

Explanation:

At x = 90 go vertical to the line for NaCl...then go left to the y-axis to find the solubility in 100 g H20 = 40

we want 300 g H20 so multiply this by 3 to get 120 gm of NaCl in 300 g

7 0

2 years ago