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brilliants [131]

3 years ago

8

An experiment is designed to test what color of light will activate a photoelectric cell the best. The photocell is set in a cir

cuit that "clicks" in response to current. The faster the current, the more clicks per minute. In this experiment, the number of clicks in one minute is recorded for each color of light shining on the photocell. To change the color of light, a different color of cellophane is placed over the same flashlight and the flashlight is then located a specific distance from the photocell.
Which of the following would be the best hypothesis for this experiment?

2 answers:

mart [117]3 years ago

8 0

Answer:hey! The answer is the color of light after it has passed through to cellophane

Explanation:

tamaranim1 [39]3 years ago

3 0

The light intensity. The transparency of the filter, the bandwidth of the filter,

Also the color temperature of light source

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The rate of decomposition of radioactive radium is proportional to the amount present at any time. The half-life of radioactive

Naya [18.7K]

Answer:

$\% \frac{A}{A_0}=75.9\%$

Explanation:

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In this case, since the kinetics of the radioactive decay is assumed to be first-order, it is possible to use the following equation to quantify that change:

$\frac{A}{A_0} =2^{-\frac{t}{t_{1/2}}$

Thus, given the elapsed time, 635 years, and the half-life, 1599 years, we can compute the fraction of the present amount:

$\frac{A}{A_0} =2^{-\frac{365years}{1599years}\\\\\frac{A}{A_0} =0.759$

Thus, the percent is:

$\% \frac{A}{A_0}=0.759*100\%\\\\ \% \frac{A}{A_0}=75.9\%$

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Answer:

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A solution of barium ions and silver ions what could be the anion

Tems11 [23]

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No two units of matter can occupy the same_______ at the same_______ .

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4 years ago

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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T

motikmotik

Answer :

(a) The rate law for the reaction is:

$\text{Rate}=k[OCl^-]^1[I^-]^1$

(b) The value of rate constant is, $60.4M^{-1}s^{-1}$

(c) rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$OCl^-+I^-\rightarrow OI^-+Cl^-$

Rate law expression for the reaction:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

where,

a = order with respect to $OCl^-$

b = order with respect to $I^-$

Expression for rate law for first observation:

$1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(1)

Expression for rate law for second observation:

$2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(2)

Expression for rate law for third observation:

$2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1$

Dividing 1 from 3, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

a = 1 and b = 1

$\text{Rate}=k[OCl^-]^1[I^-]^1$

Now, calculating the value of 'k' (rate constant) by using any expression.

$1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})$

$k=60.4M^{-1}s^{-1}$

Now we have to calculate the rate for a reaction when concentration of $OCl^-$ and $I^-$ is $1.8\times 10^{-3}M$ and $6.0\times 10^{-4}M$ respectively.

$\text{Rate}=k[OCl^-][I^-]$

$\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)$

$\text{Rate}=6.52\times 10^{-5}Ms^{-1}$

Therefore, the rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

8 0

3 years ago