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Bezzdna [24]
2 years ago
12

Carbon 14 decays to Carbon 12 and has a half-life of 5730 years. If a fossil is analysed and it has 5 grams of Carbon 14 and 15

grams of Carbon 12, how old is the fossil?
Chemistry
1 answer:
Katen [24]2 years ago
7 0

Answer:

im sorry but this question doesnt make sense

Explanation:

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A student measures a mass of an 8cm3 block of brown sugar to be 19.9G. what is the density of the brown sugar?
Naily [24]

Answer:

<h2>2.49 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question we have

density =  \frac{19.9}{8}  \\  = 2.4875

We have the final answer as

<h3>2.49 g/cm³</h3>

Hope this helps you

3 0
2 years ago
Wine has a pH of 3, which means it is __________ times more acidic than tomatoes, which have a pH of 4.
Natasha2012 [34]

Answer:

10

Explanation:

pH is defined as the negative logarithm of the concentration of hydrogen ions.

Thus,  

pH = - log [H⁺]

Thus, from the formula, more the concentration of the hydrogen ions or more the acidic the solution is, the less is the pH value of the solution.

Thus, solution with pH = 3 will be more acidic than solution with pH =4  

Thus, concentration of the [H⁺] when pH =3

3 = - log [H⁺]

[H⁺] = 10⁻³ M

For pH = 4, [H⁺] = 10⁻⁴ M

<u>hence, pH = 3 is 10 times more acidic than pH = 4</u>

5 0
3 years ago
Which type of microscope can be used to view cellular organelles such as the endoplasmic reticulum and Golgi?
Natalija [7]
Transmission electron microscope.
3 0
3 years ago
For which one of the following is ΔHfo zero?
Vlada [557]
The answer is most likely C
6 0
3 years ago
Read 2 more answers
A chemist wishing to do an experiment requiring 47-Ca2+ (half-life = 4.5 days) needs 5.0 μg of the nuclide. What mass of 47-CaCO
NARA [144]

Answer:

5.8μg

Explanation:

According to the rate or decay law:

N/N₀ = exp(-λt)------------------------------- (1)

Where N = Current quantity,  μg

            N₀ = Original quantity, μg

             λ= Decay constant day⁻¹

              t =  time in days

Since the half life is 4.5 days, we can calculate the  λ from (1) by  substituting N/N₀ = 0.5

0.5 = exp (-4.5λ)

ln 0.5  = -4.5λ

-0.6931 = -4.5λ

λ =   -0.6931 /-4.5

  =0.1540 day⁻¹

Substituting into (1)  we have :

N/N₀ = exp(-0.154t)----------------------------- (2)

To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:

N = 5.0 μg

N₀ = Unknown

t = 1 day

Substituting into (2) we have

[5/N₀]   = exp (-0.154 x 1)

    5/N₀        = 0.8572

N₀  =  5/0.8572

     =    5.8329μg

    ≈     5.8μg

The Chemist must order 5.8μg  of 47-CaCO3

6 0
2 years ago
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