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zavuch27 [327]
2 years ago
12

Carbon-14 has a half-life of 5,700 years.

Chemistry
1 answer:
Maksim231197 [3]2 years ago
5 0

Answer:

3.106\ \text{g}

Explanation:

t_{1/2} = Half-life of carbon = 5700 years

t = Time at which the remaining mass is to be found = 10400 years

m_0 = Initial mass of carbon = 11 g

Decay constant is given by

\lambda=\dfrac{\ln2}{t_{1/2}}

Amount of mass remaining is given by

m=m_0e^{-\lambda t}\\\Rightarrow m=m_0e^{-\dfrac{\ln2}{t_{1/2}} t}\\\Rightarrow m=11e^{-\dfrac{\ln 2}{5700}\times 10400}\\\Rightarrow m=3.106\ \text{g}

The amount of the substance that remains after 10400 years is 3.106\ \text{g}.

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Dima020 [189]

Answer:

Density=mass/volume

= 570/2280

=0.25g/ml

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2 years ago
Which of the following radiation sources provides the greatest radiation dose?
Sedbober [7]
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6 0
3 years ago
A 96.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 48.0 ml of koh at 25 ∘c.
PilotLPTM [1.2K]
KOH is a strong base and HBr is a strong acid and completely dissociates.
The balanced equation for the reaction is;
KOH + HBr ---> KBr + H₂O
Stoichiometry of acid to base is 1:1
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number of HBr moles reacted - 0.25 M/ 1000 mL/L x 96.0 mL  = 0.024 mol
the number of H⁺ ions are equal to number of OH⁻ ions. 
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8 0
3 years ago
A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant?
KengaRu [80]

Answer:

d. Copper (II) sulfate

Explanation:

Given data:

Mass of Al = 1.25 g

Mass of CuSO₄ = 3.28 g

What is limiting reactant = ?

Solution:

Chemical equation:

2Al + 3CuSO₄   →   Al₂ (SO₄)₃ + 3Cu

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 1.25 g/ 27 g/mol

Number of moles = 0.05 mol

Number of moles of CuSO₄:

Number of moles = mass/molar mass

Number of moles = 3.28 g/ 159.6 g/mol

Number of moles = 0.02 mol

now we will compare the moles of reactant with product.

               Al           :           Al₂ (SO₄)₃

                 2          :             1

               0.05       :          1/2×0.05=0.025 mol

                Al           :            Cu

                 2            :              3

               0.05         :            3/2×0.05 = 0.075 mol

         CuSO₄           :           Al₂ (SO₄)₃

                3             :             1

               0.02         :          1/3×0.02=0.007 mol

         CuSO₄           :            Cu

               3               :              3

               0.02         :              0.02

Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.

4 0
3 years ago
Someone plsss help me I will make you as brain plsss
Helga [31]

Answer:

2 only

Explanation:

Hope this helps

6 0
2 years ago
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