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3 years ago

I need help with my quiz.

Chemistry

2 answers:

lidiya [134]3 years ago

5 0

C because i did it lol okay bye ty so

katrin2010 [14]3 years ago

4 0

Answer:

Explanation:

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For+the+reaction+H2+++I2+-+2HI+the+equilibrium+constant,+kc+is+49+at+a+fixed+temperature.+Two+mole+of+hydrogen+and+two+moles+of+

Sonja [21]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M respectively.

Solution : Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

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

Initially 0.11 0.11 C

At equilibrium (0.11-x) (0.11-x) (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M = (0.11-x)

The expression of K_c will be,

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

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0

3 years ago

You have 4.5 x 10^24 particles of C3H8. How many grams are present?

Citrus2011 [14]

Answer:

33 g.

Explanation:

Hello there!

In this case, for these particle-mole-mass relationships problems, it is necessary for us to recall the following equivalence statement, based off the molar mass of the involved compound, C3H8, one mole of particles and the Avogadro's number:

$1mol=44.11g=6.022x10^{23}molecules$

In such a way, we can set up the following expression for the calculation of the mass in the given particles of propane:

$4.5x10^{23}molecules*\frac{1mol}{6.022x10^{23}molecules} *\frac{44.11g}{1mol}\\\\33g$

Best regards!

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3 years ago

The first pesticides relied on natural chemicals, such as those taken from chrysanthemum flowers. is this true or false

Firdavs [7]

The first pesticides relied on natural chemicals, such as those taken from chrysanthemum flowers is true

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3 years ago

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What is the volume of 88 grams of CO2 gas at STP?

I am Lyosha [343]

! mole of CO2 at STP has a volume of 22.4 liters
88 grams = 2 moles

so the required volume = 2*22.4 = 44.8 liters

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3 years ago

What is the Molar Mass

Natasha2012 [34]

The molar mass<span> is the </span>mass<span> of a chemical element or a chemical compound (g) divided by the amount of substance (mol).

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