2 years ago

Need help asap will give brainly

Chemistry

1 answer:

uranmaximum [27]2 years ago

4 0

Answer:

8. B

9. D

10. A

Explanation:

give brainliest pls ^-^

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C they are moving away from each other

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The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

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Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$

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It contains 6 sig figs

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3 years ago

5HNO3 How many atoms are in this formula. Please explain step by step.

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The coefficient 5 applis to each element in the molecule, so there are 5 times 1 H atom, 5 times 1 N atom, and ,since O has a 3 subscript, there are 5 times 3 O atoms. Therefore there are 25 atoms.

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