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UkoKoshka [18]
4 years ago
11

Three cards with holes are arranged in a straight line. A light is shined through the first card’s hole and travels through all

three cards. What does this tell you about light rays? A) that light is reflected B) that light is refractive C) that light travels in a straight line D) that light does not travel in a straight line
Chemistry
2 answers:
igomit [66]4 years ago
8 0
C) that light travels in a straight line.
Liula [17]4 years ago
7 0

Answer:

that light travels in a straight line

Explanation:

The fact that the light shone through the first card hole travels through all three cards shows that light travels in a straight line. This phenomenon is known as the rectilinear propagation of light.

If uninterrupted, light shone at one cardhole moves straight through all the cardholes in a straight line.

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The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i
Rus_ich [418]

Answer: a)  The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) concentration after 8.8 min:

8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}

\log\frac{0.25}{a-x}=0.15

\frac{0.25}{a-x}=1.41

(a-x)=0.17M

b) for concentration to decrease from 0.25M to 0.15M

t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20

t=687s

7 0
3 years ago
Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO_3)
nadya68 [22]

Answer:

CaCO₃(s) → CaO(s) + CO₂(g)

Explanation:

The decomposition reaction always make two compounds from one.

The products always have simpler chemical structure, originated from a determined compound. This can happens  spontaneously or by a third party.

A notable example of decomposition is hydrolysis. As for example the case of water, which decomposes and generates oxygen and hydrogen gas

2H₂O (l) → 2 H₂ (g) + O₂ (g)

In this case, the calium carbonate decomposes into CaO and CO₂

These two, are the products of the decomposition.

Of course, the unique reactant is the Calcium Carbonate

The balanced equation is:

CaCO₃(s) → CaO(s) + CO₂(g)

6 0
3 years ago
Read 2 more answers
How many moles of ions are produced by ionization of 2 moles of MgCl2
tensa zangetsu [6.8K]

Answer:

number of ions = 12.04 x 10^²³

Explanation:

n = number of ions/Avogadro's constant

2 = number of ions/6.02 x 10^²³

number of ions= 2 x 6.02 x 10^²³

number of ions = 12.04 x 10^²³

3 0
3 years ago
What is the maximum mass of P2I4 that can be prepared from 8.80g of P4O6 and 12.37g of iodine according to the reaction:
Dvinal [7]
The balanced chemical reaction would be as follows:

<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10

We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:

8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2

Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:

0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
8 0
3 years ago
Read 2 more answers
This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial press
Solnce55 [7]

Answer:

0.57 atm

Explanation:

When a a reaction is first order, we have from calculus the following relation:

ln[A]t/[A]₀ = - kt

where [A]t is the concentration of A ( phosphine in this case ) after a time, t

           [A]₀ is the initial concentration of A

           k is the rate constant, and

           t is the time

We also know that for a first order reaction

           k = 0.693/ t 1/2

wnere t 1/2 is the half-life.

This equation is derived for the case when A]t/= 1/2 x [A]₀ which occurs at the half-life.

Thus, lets first find k from the half life time, and then solve for t = 70.5 s

k = 0.693 /  35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

from the ideal gas law we know pV = nRT, so the volumes cancel:

ln (pPH₃ )t / p(PH₃)₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃ )t / p(PH₃)₀  = - kt

thus:

(pPH₃ )t  = 2.29 atm x e^(- 0.0198 s⁻¹ x 70.5 s ) = 0.57 atm

3 0
3 years ago
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