Answer:
Explanation:
its supposed to be not supported not inclusive because you said mix
3.0e23 atoms Ne
"E" means 10^
Then we multiply it by a mole of Ne. By the definetion of a mole, it is always 6.022e23 atoms of an element.
So now, we do this:
3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne)
After that, we use molar mass. A mole of Neon is equal, in terms of grams, to its avg. atomic mass. This goes true for any element.
It ends up like this:
3.0e23 atoms Ne x (1 mol Ne / 6.022e23 atoms Ne) x (20.1797 g Ne / 1 mol Ne)
Now cancel out the "atoms Ne" and "1 mol Ne"
You end up with a grand total of...
*plugs everything into a calculator*
10.05298... g Ne.
We need to round to 2 sig. figs. (3.0) so now it's....
10 g Ne.
Note that this method can only be used for converting atoms of an element to mass in grams.
Source(s):
A periodic table for the atomic mass of neon.
A chemistry textboook
A chemistry class.
Answer:
- <u><em>Yes, 200 ml of fluid can be transferred to a 1-quart container.</em></u>
Explanation:
You must compare the two volumes, 200 ml and 1 quart. If 200 ml is less than or equal to 1 quart, then 200 ml of fluid can be transferred to a 1-quart container, else it is not possible.
To compare, the two volumes must be on the same system of units.
Quarts is a measure of volume equivalent to 1/4 of gallon.
One gallon is approximately 3.785 liters.
3.785 liter = 3.785 liter × 1,000 ml/liter
Then, to convert 1 quart to ml use the unit cancellation method:
- (1/4)gallon × 3.785 liter/gallon × 1,000ml / liter = 946.25 ml
Thus, you get that a 1-quart container has volume of 946.25 ml, which allows that 200ml of fluid be transferred to it.
Ba2+ and Cu2+, and Sr2+ and Li+
Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole
Moles = 0.246
Volume = 280 mL = 0.280 L
Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g
Moles = 1.98
Volume = 2.6 L
Molarity = 0.762 M
Molarity = 0.76 M
