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djyliett [7]
1 year ago
6

Avanti works in a bookstore. She has four books and is going to place them in two stacks. The diagram above shows the books befo

re they touch. Use the information in the diagram to answer the question. How did the temperatures of the four books compare before they touched? How will the temperatures of the two top books compare after the books have been touching for a while, and why?
Chemistry
2 answers:
Nat2105 [25]1 year ago
6 0

Answer:

Well they didn't transfer any energy when they weren't touching and it didn't produce any energy if it didn't move. Since they are on top of each other they are causing momentum on each other creating kinetic energy.

Explanation:

Hoping this help

vivado [14]1 year ago
4 0

Answer:

cocohggghehejieieieidiiriie

Explanation:

cocobhshsh

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Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio
Paha777 [63]
N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
 The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) = 
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).

6 0
2 years ago
What is the mass of 16.3 l of helium gas?
erma4kov [3.2K]
 4.003 is the mass of helium gas


5 0
2 years ago
Suppose you take a piece of hard wax from an unlit candle. After you roll the wax between your fingers for a while, it becomes s
jolli1 [7]

Answer:

the candle is still solid..................in this case, yes!

Explanation:

it is still solid because the molecules are packed together tighter than the molecules in a liquid or gas.

6 0
2 years ago
Read 2 more answers
Compound X has the molecular formula C6H10. X decolorizes bromine in carbon tetrachloride. X also shows IR absorption at about 3
pychu [463]

Answer:

The correct answer is - option D. (check image)

Explanation:

Alkynes and alkenes both decolorized bromine in carbon tetrachloride. The absorption of the IR at about 3300 cm-1 for the X here that are found in the terminal alkynes absorption range only. In presence of excess hydrogen and a nickel catalyst, x gives the  2-methyl pentane.

The most likely structure for X is: CH3-CH3-ch-CH2-C≡CH

3 0
1 year ago
A student performed an analysis of a sample for its calcium content and obtained the following results:
sweet-ann [11.9K]

Answer:

14.9075 g, 28.67%, 0.11%

Explanation:

The mean concentration of calcium = summation x / frequency

= ( 14.92 + 1491 + 14.88 + 14.92 ) /4 = 14.9075 g

Standard deviation = √(summation (x - μ)² /n) = √ ( ((14.92 - 14.9075)² +(14.91 - 14.9075)² + (14.88 - 14.9075)² + ( 14.92 - 14.9075)²) / 4) = 0.0164

b)  percent error = abs(14.9075 - 20.90) / 20.90 × 100 = 28.67%

c) relative standard deviation = standard deviation / mean × 100 = 0.0164 /  14.9075 × 100 = 0.11%

d) The accuracy of the measure is the measurement compared to the actual which according to the standard set by the instructor (5%error) is not very accurate because the percent error is high (28.67%) while the relative standard deviation is quite low ( 0.11%) which means the measurement precision is very high.

The student will have to redo the experiment because the experiment was not too accurate since the percent error is way higher than the set value (5%) although the precision was high.

5 0
2 years ago
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