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kogti [31]
3 years ago
8

A geologist is attempting to date a layer of sedimentary rock. He performs radioactive dating using uranium-235, which has a hal

f-life of 703 million years. He determines that the uranium-235 has undergone 3 half-lives. Approximately how old is the rock?
710 billion years

2.11 billion years

710 million years

235 million years

2.11 million years

none of the above
Chemistry
2 answers:
Sindrei [870]3 years ago
6 0
It’s 2.11 billion years
Bad White [126]3 years ago
4 0

Answer:

2.11 billion years.

Explanation:

I hope this helps! Have a great rest of your day!

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The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.
Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

Start with 0.0020 moles of iodate ions {\text{IO}_{3}}^{-}.

How many moles of iodine \text{I}_2 will be produced?

{\text{IO}_{3}}^{-} converts to \text{I}_2 in the first reaction. The coefficient in front of \text{I}_2 is three times the coefficient in front of {\text{IO}_{3}}^{-}. In other words, each mole of {\text{IO}_{3}}^{-} will produce three moles of \text{I}_2. 0.0020 moles of {\text{IO}_{3}}^{-} will convert to 0.0060 moles of \text{I}_2.

How many moles of thiosulfate ions {\text{S}_2\text{O}_3}^{2-} are required?

\text{I}_2 reacts with {\text{S}_2\text{O}_3}^{2-} in the second reaction. The coefficient in front of \text{I}_2 is twice the coefficient in front of {\text{S}_2\text{O}_3}^{2-}. How many moles of {\text{S}_2\text{O}_3}^{2-} does each mole of \text{I}_2 consume? Two. 0.0060 moles of \text{I}_2 will be produced. As a result, 2 \times 0.0060 = 0.0120 moles of {\text{S}_2\text{O}_3}^{2-} will be needed.

6 0
3 years ago
The stored energy in a battery can BEST be described as:
34kurt
Chemical is the answer your looking for.
3 0
3 years ago
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Calculate a reasonable amount (mass in g) of your unknown acid to use for a titration. You will want about 30 mL of titrant to g
Vinil7 [7]

Answer:

"0.60 g" is the appropriate solution.

Explanation:

The given values are:

Volume of base,

= 30 ml

Molarity of base,

= 0.05 m

Molar mass of acid,

= 400 g/mol

As we know,

⇒  Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}

On substituting the values, we get

⇒           0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}

⇒  Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}

⇒                                             =1.5\times 10^{-3}  

hence,

⇒  Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}

On substituting the values, we get

⇒  1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}

⇒  Mass \ of \ acid=1.5\times 10^{-3}\times 400

⇒                         =0.60 \ g

8 0
3 years ago
Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y
olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

\lambda =\frac{0.693}{t_{\frac{1}{2}}}

A=A_o\times e^{-\lambda t}

Where:

\lambda= decay constant

A_o =concentration left after time t

t_{\frac{1}{2}} = Half life of the sample

Half life of Pu-239 = t_{\frac{1}{2}}=24,000 y[

\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]

Let us say amount present of  Pu-239 today = A_o=x

A = ?

A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}

A=0.9715\times x

\frac{A}{A_0}=\frac{A}{x}=0.9715

0.9715 Fraction of Pu-239 will be remain after 1000 years.

7 0
3 years ago
Calculate total ATP produced from a fatty acid of 32 carbons
emmasim [6.3K]

Answer:

Total number of ATP molecules generated from a 32-carbon fatty acid = 206 ATP molecules

Explanation:

A 32 carbon fatty acid which undergoes complete beta-oxidation assuming that the fatty acid is fully saturated will pass through the beta-oxidation cycle 14 times to produce the following:

15 molecules of acetylCoA, 14 molecules of FADH₂, and 14 molecules of NADH.

Each of the 15 acetylCoA molecules can be further oxidized in the citric acid cycle to yield the following: 15 × 3 NADH; 15 × 1 FADH₂, and 15 ATP molecules from the substrate level phosphorylation occuring at the succinylCoA synthetase catalyzed-reaction.

Total FADH₂ produced = 15 + 14 = 29 molecules of FADH₂

Total NADH produced = 45 + 14 = 59 molecules of NADH

The FADH₂ and NADH will each donate a pair of electrons to the electron transfer flavoprotein and mitochondrial NADH dehydrogenase respectively of the electron transport chain, and about 1.5 and 2.5 molecules of ATP are generated respectively when these electrons are transfered to molecular oxygen.

Thus, number of molecules of ATP generated by 29 molecules of FADH₂ = 1.5 × 29 = 43.5 molecules of ATP.

Number of molecules of ATP generated by 59 molecules of NADH = 2.5 × 59 = 147.5

Sum of ATP generated from FADH₂ and NADH = 43.5 + 147.5 = 191 ATP molecules

Total number of ATP molecules generated = 191 + 15 = 206 ATP molecules

Total number of ATP molecules generated from a 32-carbon fatty acid = 206 ATP molecules

7 0
3 years ago
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