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3 years ago

How much of a 0.74mg sample of uranium-235 will remain after 2.8x10^9

1 answer:

5 0

Of Uranium-235 remains after 2.8 x 10^9 years, what was the original mass of the sample of Uranium-235? The half-life of Uranium-235 is 7.0 x 10^8 years. Uranium-232 has a half life of 68.8 years.

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___∆ 1. Erosion is / is not caused by water, ice, wind, chemicals, and sand. (pick one)

svlad2 [7]

Answer:

1. Chemicals

2. Erosion

3. Solids

4. Expands

5. Chemical

6. Arch

7. Sand

8. Surface

Explanation:

7 0

3 years ago

In the periodic table, how are elements in the same group related to each other? A. They have the same atomic mass. B. They have

Irina-Kira [14]

Elements in the same group tend to have very similar properties (D). This is due to the number of valence electrons each group has.

3 0

3 years ago

Read 2 more answers

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

(a) a solution of pH 3.0

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

(b) a solution of 0.10 M $NH_3$

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

(c) a solution with a pOH of 12.

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

(d) a solution of 0.10 M NaOH

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

(e) a $1\times 10^{-4}M$ solution of $HNO_2$

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

3 years ago

10decimeter= 10Decameter=

snow_lady [41]

Answer:

Explanation:

No, I don't owe nothing

Stop acting funny 5I've seen them running

Coming for more

4646

3 0

3 years ago

What is the balanced form of the chemical equation shown below?

anzhelika [568]

Answer:

C2H5OH +3O2 = 2Co2+ 3H2O

Explanation:

3 0

3 years ago