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Ratling [72]
3 years ago
7

​If a bag of peas weighs 454 grams, how many peas would be in the bag?

Chemistry
1 answer:
OverLord2011 [107]3 years ago
6 0

​In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

The average pea weighs between 0.1 and 0.36 grams.

If we take the lower value (0.1 g/pea), the number of peas in 454 g is:

454 g \times \frac{1pea}{0.1 g} = 4540pea

If we take the higher value (0.36 g/pea), the number of peas in 454 g is:

454 g \times \frac{1pea}{0.36 g} \approx  1261pea

​In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

You can learn more about conversion factors here: brainly.com/question/1844638

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Answer:

The volume of the gasoline in the tank is 18.85 cubic feet

Explanation:

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I hope my answer helps you

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Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c
uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

I_{2}  (g) + Cl_{2}  (g) ----->  2 ICl (g)

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

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Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = \frac{(2x)^{2}}{(0.437-x) (0.269-x)}

81.9 = \frac{(2x)^{2}}{x^{2} -0.706x + 0.118}

(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]

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Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

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Therefore the change in concentration of the gases during the reaction is 0.254 M

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[ICl]_{eq} = 2 ( 0.254) = 0.508 M

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