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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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G. Amount of charge required to reduce
Nady [450]

Answer:

\boxed{\text{c) 4 F}}

Explanation:

1. Write the skeleton equation for the half-reaction

NO₃⁻ ⟶ N₂O

2. Balance all atoms other than H and O

2NO₃⁻ ⟶ N₂O

3. Balance O by adding H₂O molecules to the deficient side.

2NO₃⁻ ⟶ N₂O + 5H₂O

4. Balance H by adding H⁺ ions to the deficient side.

2NO₃⁻ + 10H⁺ ⟶ N₂O + 5H₂O

5. Balance charge by adding electrons to the deficient side.

2NO₃⁻ + 10H⁺ + 8e⁻ ⟶ N₂O + 5H₂O

The amount of charge required to reduce 2 mol of NO₃⁻ is 8 F

\text{The amount of charge required to reduce 1 mol of NO$_{3}^{-}$ is \boxed{\textbf{4 F}}}

4 0
2 years ago
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A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0
3 years ago
Which choice is not an example of a molecule? OF O H202 O 03 O NC13​
MissTica

Answer:

Think it's NC13

Explanation:

It's the only one missing in the molecule

5 0
3 years ago
Suppose that on a dry, sunny day when the air temperature is near 37 ∘C,37 ∘C, a certain swimming pool would increase in tempera
ioda

Answer:

The fraction of water body necessary to keep the temperature constant is 0,0051.

Explanation:

Heat:

Q=m*Ce*ΔT

Q= heat  (unknown)

m= mass  (unknown)

Ce= especific heat (1 cal/g*°C)

ΔT= variation of temperature  (2.75 °C)  

Latent heat:

ΔE=∝mΔHvap

ΔE= latent heat

m= mass  (unknown)

∝= mass fraction (unknown)

ΔHvap= enthalpy of vaporization (539.4 cal/g)

Since Q and E are equal, we can match both equations:

m*Ce*ΔT=∝*m*ΔHvap

Mass fraction is:

∝=\frac{Ce*ΔT}{ΔHvap}

∝=\frac{(1 cal/g*°C)*2.75°C}{539.4 cal/g}

∝=0,0051

7 0
2 years ago
How are the electron structures of boron (B) and aluminum (Al) similar?
andrezito [222]
The electron structures of boron and aluminum are similar because they share the same group, therefore they have the same amount of valence electron. 
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