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Vesna [10]
3 years ago
10

If you have 100 grams of a radioactive isotope with a half life of 10 years how much of the isotope will you have left after 20

years
Chemistry
1 answer:
Cerrena [4.2K]3 years ago
3 0

Hey there!

A half-life means after a certain amount of time, half of that substance will be gone/changed after that time.

There are two half lives in 20 years because 20 ÷ 10 = 2.

So, we divide the 100g sample in half 2 times.

100 ÷ 2 = 50

50 ÷ 2 = 25

There will be 25g of the radioactive sample remaining after two half lives.

Hope this helps!

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Which compounds are classified as Arrhenius acids?
Olenka [21]

Answer:The correct answer is option 4.

Explanation:

Arrhenius acids are those compounds which gives H^+ ions when dissolved in their aqueous solution.

HA(aq)\rightarrow H^++A^-

Arrhenius bases are those compounds which gives OH^- ions when dissolved in their aqueous solution.

BOH(aq)\rightarrow OH^-+B^+

HBr \& H_2SO_4 are Arrhenius acids because they form H^+ions in their respective aqueous solution.

HBr(aq)\rightarrow H^++Br^-

H_2SO_4(aq)\rightarrow 2H^++SO_{4}^{2-}

Hence, the correct answer is option 4.

6 0
3 years ago
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State the difference between sugar in water and pebbles in water
Usimov [2.4K]
Sugar is made of molecules that are bonded together based on the positively and negatively charged areas.  They will slowly dissolve in water.  Pebbles are solids.  They will sit in water for a long time.  Though shale pebbles will break apart or fall apart.
6 0
3 years ago
Calculate the approximate volume of a 0.6000mol sample of gas at 288.15K and a pressure of 1.10atm.
11111nata11111 [884]

Answer:

The volume of the sample of the gas is found to be 12.90 L.

Explanation:

Given pressure of the gas = P = 1.10 atm

Number of moles of gas = n = 0.6000 mole

Temperature = T = 288.15 K

Assuming the volume of the gas to be V liters

The ideal gas equation is shown below

\textrm{PV} =\textrm{nRT} \\1.10 \textrm{ atm}\times V \textrm{ L} = 0.6000 \textrm{ mole}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 288.5\textrm{K} \\\textrm{V} = 12.90 \textrm{ L}

Volume occupied by gas = 12.90 L

6 0
4 years ago
10.00 mL of 0.10 M NaOH is added to 25.00 mL of 0.10 M HCl during a titration
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Answer:

Explanation:

10 mL = .01 L .

25 mL = .025 mL .

10 mL of .1 M NaOH will contain .01 x .1 = .001 moles

25 mL of .1M HCl will contain .025 x .1 = .0025 moles

acid will neutralise and after neutralisation moles of acid remaining

= .0025 - .001 = .0015 moles .

Total volume = .01 + .025 = .035 L

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Option D is correct.

= .042857 M

= 42.857 x 10⁻³ M .

pH = - log [42.857 x 10⁻³]

= 3 - log 42.857

= 3 - 1.632

= 1.368 .

3 0
3 years ago
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