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vekshin1
3 years ago
14

Carbon-14 has a half-life of approximately 5700 years. How much of a 1000g sample will be left undecayed and still radioactive a

fter 11,400 years?
A: 1000g
B: 500g
C: 250g
D: 125g
Chemistry
1 answer:
Artist 52 [7]3 years ago
4 0

Answer:

option C is correct (250 g)

Explanation:

Given data:

Half life of carbon-14 = 5700 years

Total amount of sample = 1000 g

Sample left after 11,400 years = ?

Solution:

First of all we will calculate the number of half lives passes during 11,400 years.

Number of half lives = time elapsed/ half life

Number of half lives = 11,400 years/5700 years

Number of half lives = 2

Now we will calculate the amount left.

At time zero = 1000 g

At first half life = 1000 g/2 = 500 g

At second half life = 500 g/2 = 250 g

Thus, option C is correct.

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24.5872 meters per second

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3 years ago
Atomic Size: Determine which of the following element pairs is the larger one
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Answer: I think the answer is 3.) Krypton or Argon.

Explanation:

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3 years ago
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 3.30kg of water at 23.8°C. Dur
Brums [2.3K]

Answer:

T_2= 31.9\°C

Explanation:

Hello there!

In this case, it is possible to propose an energy balance in order to illustrate how the heat released by the reaction is absorbed by the water:

-Q_{rxn}=Q_{water}

Thus, since the heat released by the reaction is -112 kJ (-112000 J), it is possible to define the hear absorbed by the water in terms of mass, specific heat and temperature change:

-(-112000J)=m_{water}C_{water}(T_2-T_1)

In such a way, it is possible to define the final temperature as shown below:

T_2=23.8\°C+\frac{112000J}{3300g*4.18\frac{J}{g\°C} }\\\\T_2= 31.9\°C

Best regards!

4 0
3 years ago
A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin
lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, 3.0\times 10^1g

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron metal = 0.45J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of iron metal = 32.3 g

m_2 = mass of water = ?

T_f = final temperature of mixture = 59.2^oC

T_1 = initial temperature of iron metal = 21.9^oC

T_2 = initial temperature of water = 63.5^oC

Now put all the given values in the above formula, we get

32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC

m_2=30.16g\approx 3.0\times 10^1g

Therefore, the mass of the water in two significant figures is, 3.0\times 10^1g

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3 years ago
What is the net force on this box?<br> A.) 0<br><br> B.) 45<br><br> C.)25<br><br> D.)90
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Answer:

D

Explanation:

Hopefully this helps you!

3 0
3 years ago
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