Answer:
28500 years
Explanation:
Applying,
A = A'(
)............... Equation 1
Where A = Original mass of Carbon-14, A' = Final mass of carbon-14 after decaying, x = total time, y = half-life.
From the question,
Given: A = 1 g, A' = 31.3 mg = 0.0313 g, y = 5700 years.
Substitute these values into equation 1
1 = 0.0313(
)
= 1/0.0313
= 31.95
≈ 32
≈ 2⁵
Equating the base and solve for x
x/5700 ≈ 5
x ≈ 5×5700
x ≈ 28500 years
Explanation:
The Lewis dot diagram shows how electrons participate in a bond with Carbon and Chlorine. This is shown by the sticks and the 2 paired electrons near the carbon atom which represent the bonds. These electrons form these bonds because they form octets when they are bonded which most molecules and compounds follow
Hoped this helped, 2Trash4U
<u>36 ml of NaOh and</u><u> 464 ml</u><u> of </u><u>HCOOH</u><u> would be enough to form 500 ml of a buffer with the same pH as the buffer made with </u><u>benzoic acid </u><u>and NaOH.</u>
What is benzoic acid found in?
- Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
- Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.
Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.
Amount of moles of NaOH -2 × 0.025 = 0.05 mol
Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol
In this case, we can calculate the pH produced by the buffer of these two reagents, as follows


We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows




Now we must solve the equation above. This will be done using the following values

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.
NaOH volume
( 0.5 - 0.464)L
0.036L .................... 36ml
HCOOH volume
500 - 36 = 464mL
Learn more about benzoic acid
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Answer:
68133080.02 g
Explanation:
I believe that the question is to find the mass of air in the room and not the molar mass of air since the molar mass of air was already given in the question as 28.97 g/mol.
Now, if 1 mole of a gas occupies 22.4 L
x moles of air occupies 52,681,428.8 Liters
x = 1 * 52,681,428.8 /22.4
x = 2351849.5 moles of air
Now, number of moles = mass/ molar mass
but molar mass = 28.97 g/mol
2351849.5 = mass/28.97
mass = 2351849.5 * 28.97
mass = 68133080.02 g