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lora16 [44]
2 years ago
5

6 The half-life of carbon-14 is 5,700 years. A sample of wood that originally contained 100 grams of carbon-14 now contains only

25 grams of carbon-14. Approximately how many years ago was this sample part of a living tree? * (6 Points)​
Chemistry
1 answer:
nikitadnepr [17]2 years ago
8 0

Answer:

The correct answer is - 10400 years.

Explanation:

Half-life an element is the time required to decay the half amount of the initial amount of the element.

Here, the initial amount n(i) = 100 gm

remained amount n(r) = 25 gm

then n =?

The formula for the calculating initial amount

n(i)=n(r)⋅2^n

n(i)=n(r)⋅2^n

n(i)/n(r) =2^n

puting value,

100/25 = 2^n

2^n = 4

n = 2

so to decay this amount there is two half life cycles involved,

Thus, time = 2*5700

= 10400 years.

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2 years ago
A 2.00-mol sample of hydrogen gas is heated at constant pressure from 294 K to 414 K. (a) Calculate the energy transferred to th
Furkat [3]

Answer:

a) The energy transferred is 6.91 kJ

b) The internal energy is 4.90 kJ

c) The work done on the gas is - 2.01 kJ

Explanation:

Step 1: Data given

Number of moles of hydrogen gas = 2.00 moles

Pressure = constant

Temperature is heated from 294 K to 414 K

Molar heat capacity of hydrogen gas = 28.8 J/mol*K

Step 2: Calculate the energy transferred to the gas by heat.

Q = n* Cp * ΔT

⇒with Q =the energy transferred

⇒with n = the number of moles = 2.00 moles

⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K

⇒ with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

Q = 2.00 * 28.8 * 120

Q = 6912 J = 6.91 kJ

Step 3: Calculate the increase in its internal energy.

ΔEint = n*Cv*ΔT

⇒with ΔEint = the increase in its internal energy.

⇒with n = the number of moles = 2.00 moles

⇒with Cv = The constant volume = 20.4 J/mol*K

⇒with  ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

ΔEint = 2.00 * 20.4 * 120

ΔEint =4896 J = 4.90 kJ

Step 4: Calculate the work done on the gas.

Work done on the gas = -Q + ΔEint

W = -6.91 kJ + 4.90 kJ

W = -2.01 kJ

6 0
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Read 2 more answers
Calculate the number of grams of solute in 814.2mL of 0.227 M calcium acetate
kiruha [24]

Answer:

Mass = 29.23 g

Explanation:

Given data:

Volume of solution = 814.2 mL 814.2/1000 = 0.8142 L)

Molarity of solution = 0.227 M

Mass of solute in gram = ?

Solution:

Molarity = number of moles / volume in L

By putting values,

0.227 M = number of moles / 0.8142 L

Number of moles = 0.227 M × 0.8142 L

Number of moles = 0.184 mol

Mass in gram:

Mass = number of moles × molar mass

Molar mass of calcium acetate = 158.17 g/mol

Mass = 0.184 mol × 158.17 g/mol

Mass = 29.23 g

6 0
3 years ago
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