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Temka [501]
2 years ago
6

How do you rewrite 4 5/7 as an improper fraction. in steps

Chemistry
2 answers:
Tom [10]2 years ago
7 0

Answer:

33/7

Explanation:

1. take the fraction and multiply the denominator by 4 to get 28

2. add the numerator to that... 28+5=32

noname [10]2 years ago
6 0

Answer:

33/7

Multiply the denominator of the fraction by the whole number.

Add this result to the numerator of the fraction

answer becomes the numerator of the improper fraction.

Explanation:

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What current (in a) is required to plate out 1. 22 g of nickel from a solution of ni2 in 0. 50 hour?
Mashcka [7]

The current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.

<h3>What is current?</h3>

The current is given as the product of the charge with time. In the electrochemical analysis of the nickel, there will be a reduction of the nickel ion to nickel. The formation is given as:

\rm Ni^{2+}+2e^-\;\rightleftharpoons Ni

There is the deposition of 1 mole of Ni with 2 electrons transfer. The transfer of charge for 1 mole that is 58.7 grams Nickel is:

\rm 58.7\;g=2\;\times\;96487\;C\\58.7\;g=192974\;C

The mass of Ni to be deposited is 1.22 grams. The charge required is given as:

\rm 58.7\;grams\;Ni=192974\;C\\\\1.22\;grams\;Ni=\dfrac{192974}{58.7}\;\times\;1.22\;C\\\\1.22\;grams\;Ni=4010.7\;C

The current required to transfer 4010.7 C of charge in 1800 seconds is given as:

\rm Charge=Current\;\times\;Time\\4010.7\;C=Current\;\times\;1800\;sec\\Current=2.23\;A

Thus, the current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.

Learn more about current, here:

brainly.com/question/23063355

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7 0
2 years ago
The half-life for the radioactive decay of C-14C-14 is 5730 years. You may want to reference (Pages 598 - 605) Section 14.5 whil
KIM [24]

<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life of the reaction = 5730 years

Putting values in above equation, we get:

k=\frac{0.693}{5730yrs}=1.21\times 10^{-4}yrs^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

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k = rate constant  = 1.21\times 10^{-4}yr^{-1}

t = time taken for decay process = ? yr

[A_o] = initial amount of the sample = 100 grams

[A] = amount left after decay process =  (100 - 25) = 75 grams

Putting values in above equation, we get:

1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{75}\\\\t=2377.9yrs

Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

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