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3 years ago

Question:

Chemistry

1 answer:

anzhelika [568]3 years ago

7 0

Answer:

A. Chemical

B. physical

Explanation:

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Question 3

svp [43]

I added a screenshot because I recently answered the same question. I hope this helps you!! :)

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2 years ago

The radioisotope that has the longest half-life is the best to use in powering planet and space exploration vehicles because the

Diano4ka-milaya [45]

Answer:

U-238

Explanation:

For a given radioisotope, half life can be defined as the time taken for the isotope to decay into one-half of its original amount. Mathematically, this is expressed as:

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

where k = rate constant for the radioactive decay process

Greater the t1/2 longer will be its stability.

Based on the given data, U-238 has the largest half life and therefore will be best suited for applications mentioned.

4 0

3 years ago

Read 2 more answers

Scientific Notation - PLEASE HELP!

zmey [24]

Answer:

0.0931 is the ans i think

8 0

4 years ago

A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o

statuscvo [17]

Answer: The value of $K_{eq}$ is 0.044

Explanation:

We are given:

Initial moles of methane = $4.10\times 10^{-2}mol=0.0410moles$

Initial moles of carbon tetrachloride = $6.51\times 10^{-2}mol=0.0651moles$

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of methane = $\frac{0.0410}{1.00}=0.0410M$

Concentration of carbon tetrachloride = $\frac{0.0651}{1.00}=0.0651M$

The given chemical equation follows:

$CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)$

Initial: 0.0410 0.0651

At eqllm: 0.0410-x 0.0651-x 2x

We are given:

Equilibrium concentration of carbon tetrachloride = $6.02\times 10^{-2}M=0.0602M$

Evaluating the value of 'x', we get:

$\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M$

Now, equilibrium concentration of methane = $0.0410-x=[0.0410-0.0049]=0.0361M$

Equilibrium concentration of $CH_2Cl_2=2x=[2\times 0.0049]=0.0098M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044$

Hence, the value of $K_{eq}$ is 0.044

8 0

3 years ago

Why were chlorofluorocarbons first developed?

Sergio [31]

CFCs and their associated compounds were developed in the early 1900s as a non-toxic, non-flammable solution to other more dangerous products such as ammonia.

5 0

4 years ago