Molar mass of C: 12.011 g/mol
The equation says C20, which means there are 20 carbon atoms in each molecule of Vitamin A. So, we multiply 12.011 by 20 to get 240.22 g/mol carbon.
Molar mass of H: 1.0079 g/mol
The equation says C30, which means there are 30 hydrogen atoms in each molecule of Vitamin A. So, we multiply 1.0079 by 30 to get 30.237 g/mol hydrogen.
Molar mass of O: 15.999 g/mol
The equation says O without a number, which means there is only one oxygen atom in each molecule of Vitamin A. So, we leave O at 15.999 g/mol.
Then, just add it up:
240.22 g/mol C + 30.237 g/mol H + 15.999 g/mol O = 286.456 g/mol C20H30O
So, the molar mass of Vitamin A, C20H30O, is approximately 286.5 g/mol.
Answer:
The beaker holds 307.94 mL
Explanation:
As we know that the volume that beaker hold is the volume of water that occupied by it.
For this first we have to find mass of the water in the beaker
This can be calculated by the subtraction of beaker's weight from the weight of beaker and water.
weight of water (m) = total weight - weight of beaker
Empty weight of beaker = 25.91 g
Weight of beaker with water = 333.85 g
Weight of water = 333.85 - 25.91 = 307.94 g
Density of water = 1 g/mL
We have
Mass = Volume x density
307.94 = Volume x 1
Volume = 307.94 mL
The beaker holds 307.94 mL
It has a negative charge emits energy when it moves to a lower energy orbit from an excited state and it has the same mass as a proton
Answer:
There are 0.219 mol of LINO3
Answer:
The correct answer is 0.12 grams.
Explanation:
The mass of carbon monoxide or CO collected in the tube can be determined by using the ideal gas equation, that is, PV = nRT.
Based on the given question, P or the pressure of the gas is given as 1 atm, volume of the gas collected in the tube is 117 ml or 0.117 L.
The number of moles or n can be determined by using the equation, mass/molar mass.
R is the universal gas constant, whose value is 0.0821 L atmK^-1mol^-1, and temperature is 55 degree C or 328 K (55+273).
On putting the values we get:
n = PV/RT
= (1 atm*0.117 L) / (0.0821 L atmK^-1mol^-1 * 328 K)
= 0.0043447 mol
Therefore, mass of CO will be moles * molar mass of CO
= 0.0043447 mol * 28 g/mol
= 0.12 g
