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mario62 [17]
3 years ago
5

How does vibrational energy contribute to a reaction?

Chemistry
1 answer:
Viefleur [7K]3 years ago
6 0

Answer:

A lot of thing tbh

Explanation:

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How does an atom-88 become a strontium ion with a +2 charge?
Fantom [35]

Strontium atom loses 2 electrons to become an ion with 2 electrons lesser than its atom. Your answer is C.

6 0
3 years ago
Write the number 345 in scientific notation, with the same number of significant figures:
uranmaximum [27]

Answer:

3.45 x 10^2

Explanation:

345= 3 SF

3.45 x 10^2= 3 SF

8 0
3 years ago
Which bones are shortened in brachycephalic dogs?
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Bones affected in brachycephalic dogs are the jaw bones, and part of the skull.
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4 years ago
Read 2 more answers
A cat travels 240 meters in 30 seconds, and a sprinter travels 100 meters in 9.5 seconds. Which is traveling faster, the cat or
joja [24]

Answer:

The cat

Explanation:

1) You need to find how many meters per second each thing is sprinting, so you can divide the meters by seconds for each.

\frac{240}{30} = 8 m/s\\\frac{100}{9.5} = around  10.5 m/s

2) Per second, the sprinter is able to achieve 8 meters, while the cat can do around 10.5 meters. 10.5 is greater than 8, making the cat faster than the sprinter.

4 0
2 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
4 years ago
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