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JulsSmile [24]
3 years ago
14

6. A group of students is performing a simple scientific investigation with baking soda and vinegar. The students place vinegar

in a beaker and put baking soda inside of a balloon. Then, the balloon with the baking soda is attached to the mouth of the beaker. The reaction of the baking soda being poured from the balloon into the beaker is shown Baking soda falls Vinegar and baking soda mixture into vinegar What evidence shows that a chemical change occurred with the baking soda and vinegar?
A The vinegar and baking soda released a gas.

B. The vinegar and baking soda gave off an odor.

C. The vinegar and baking soda produced a flame

. D. The vinegar and baking soda changed in temperature.​
Chemistry
1 answer:
natali 33 [55]3 years ago
5 0

Answer:

A. The vinegar and baking soda released a gas. Hope it helps. :)

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Which of the following is a correct formula unit? PbCl AlP Li3Al ClO
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Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
Katen [24]

<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

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

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

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