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3 years ago

A radioactive element is half disintegrated in 40 minutes. What is the time required for the decay of 75% of the element

Chemistry

1 answer:

natali 33 [55]3 years ago

8 0

80 minutes its half life is 40 mins and another 40 would decay half of the 50% left currently

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How many moles are in 2.54 * 10^73 atoms of copper?

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Answer:

= 3.99 ⋅ m o l ⋅ copper atoms ...a mass of approx. 250 ⋅ g

Explanation:

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3 years ago

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3 years ago

Some COCl2 is placed in a sealed flask and heated to 756 K. When equilibrium is reached, the flask is found to contain COCl2 (7.

o-na [289]

Answer:

$9.044\times 10^{-3}$ is the value of the equilibrium constant for this reaction at 756 K.

Explanation:

$COCl_2\rightleftharpoons CO+Cl_2$

Equilibrium concentration of $COCl_2$

$[COCl_2]=7.40\times 10^{-4} M$

Equilibrium concentration of $CO$

$[CO]=3.76\times 10^{-2} M$

Equilibrium concentration of $Cl_2$

$[Cl_2]=1.78\times 10^{-4} M$

The expression of an equilibrium constant can be written as;

$K_c=\frac{[CO][Cl_2]}{[COCl_2]}$

$=\frac{3.76\times 10^{-2}\times 1.78\times 10^{-4}}{7.40\times 10^{-4}}$

$K_c=9.044\times 10^{-3}$

$9.044\times 10^{-3}$ is the value of the equilibrium constant for this reaction at 756 K.

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3 years ago

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Trava [24]

Answer:

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Explanation:

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2 years ago

Given: 2LiBr + I2 → 2LiI + Br2 Calculate the mass of bromine produced when 9.033 × 1023 particles of iodine (I2) react completel

Llana [10]

Answer:

239.7 g

Explanation:

Step 1: Write the balanced equation

2 LiBr + I₂ → 2 LiI + Br₂

Step 2: Convert the molecules of iodine to moles

We have 9.033 × 10²³ particles (molecules) of iodine. In order to convert molecules to moles, we will use the Avogadro's number: there are 6.022 × 10²³ molecules of iodine in 1 mole of iodine.

$9.033 \times 10^{23}molecule \times \frac{1mol}{6.022 \times 10^{23}molecule} =1.500mol$

Step 3: Calculate the moles of bromine produced

The molar ratio of I₂ to Br₂ is 1:1. Then, the moles of bromine produced are 1.500 moles.

Step 4: Calculate the mass of bromine

The molar mass of bromine is 159.81 g/mol. The mass corresponding to 1.500 moles is:

$1.500mol \times \frac{159.81g}{mol} = 239.7 g$

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3 years ago