The volume of a 1.86-carat diamond in cubic centimeters is 0.106 cm³
Given,
The density of a diamond is 3.513 g/cm³.
We have to find out the volume of a 1.86-carat diamond in cubic centimeters.
Convert the units of the diamond from carat to grams, we have:
(1.86 carats) x (0.200 g / 1 carat) = 0.372 g
The volume of the diamond is obtained by dividing the mass by the density, therefore using the formula, we get
v = m / d
v = 0.372 g / (3.51 g/cm³) = 0.1059 cm³
or, v = 0.106 cm³ (approx)
Therefore, the volume of a 1.86-carat diamond is approximately 0.106 cm³.
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The formula for pH given the pKa and the concentrations
are:
pH = pKa + log [a–]/[ha]
<span>
Therefore calculating:</span>
3.75 = 3.75 + log [a–]/[ha]
log [a–]/[ha] = 0
[a–]/[ha] = 10^0
<span>[a–]/[ha] = 1</span>
Explanation:
Sorry, I don't know, but I can tell you that when an atom, or a body, has the same amount of positive charges (protons) and negative charges (electrons), it is said to be electrically neutral. ... The net charge corresponds to the algebraic sum of all the charges that a body possesses.
Answer:
There are 1000 mg in 1. g
There are 1000 g in 1 kg
Each students needs 2,250 mg of clay
Explanation:
In order to determine the amount of how many kilograms to order, the teacher will need to find out the total mass of clay required by the students and then the teacher will have to convert the total mass to the units required for the purchase.
The teacher would have to find out how many milligrams make one kilogram as follows
1 kg = 1000 g
1 g = 1000 mg
Therefore, 2,250 mg = 2250/1000 g = 2.25 g
2.25 g = 2.25/1000 kg = 0.00225 kg.
<u>Answer:</u>
The correct answer option is a. Selenium (Se).
<u>Explanation:</u>
From the given answer options, Selenium (Se) is the only element which is a non metal.
It has the atomic number 34 with atomic weight 78.96. It is a member of the sulfur group of the non metallic elements and falls in period 4 of the Periodic table.
Selenium has non metallic properties which are intermediate between the elements that lie above and below it in the Periodic table.
