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grandymaker [24]
3 years ago
12

Use the graph to determine the velocity between 6 and 12 seconds. m/s What was the velocity over the entire trip? m/s

Chemistry
2 answers:
likoan [24]3 years ago
6 0

Answer:

-2 m/s

0 m/s

Explanation:

I just did it

natka813 [3]3 years ago
6 0

<u>Answer:</u> The velocity of the particle is -2 m/s

<u>Explanation:</u>

Velocity is defined as the ratio of displacement of a particle to the time taken by the particle. The units of velocity is meters per second.

The equation representing velocity of a particle is given as:

\text{Velocity}=\frac{\text{Displacement}}{\text{Time}}

From the graph:

The particle has traveled a distance of 12 m in the downward direction in between 6 to 12 seconds.

Displacement of particle = -12 m

Time taken = 12 - 6 = 6 sec

Putting values in above equation, we get:

\text{Velocity}=\frac{-12m}{6sec}=-2m/s

Negative side represents the particle is moving in downward direction.

Hence, the velocity of the particle is -2 m/s

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Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

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Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

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[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

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viva [34]

Answer:

8.934 g

Step-by-step explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:        192.12                                                                   44.01

            H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂

m/g:        13.00

For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.

(a) Calculate the <em>moles of H₃Cit </em>

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n = 0.067 67 mol H₃Cit

(b) Calculate the <em>moles of CO₂ </em>

The molar ratio is (3 mol CO₂/1 mol H₃Cit)

n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)

n = 0.2030 mol CO₂

(c) Calculate the <em>mass of CO₂ </em>

m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)

m = 8.934 g CO₂

4 0
3 years ago
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