<span>1.
</span>The balanced reaction is:<span>
Mg (s) + 2HCl (aq)
--> MgCl 2 (aq) + H 2 (g)
We
are given the amount hydrochloric acid to be used for the reaction. This will
be the starting point of the calculation.
40.0 g
HCl ( 1 mol HCl / 36.46 g HCl)
(1 mol H2 / 2 mol HCl) (2.02 g H2 / 1 mol H2) = 1.11 g H2</span>
Answer:
D. All of the above
Explanation:
We can place a piece of glass tubing into rubber stopper after the tubing has been fire polished and cooled by
Lubricating the tubing with water or glycerin.
Using a towel or cotton gloves for protection.
or by
Twisting the tubing and stopper carefully.
Hence we can say that all the above methods are equally useful for the above procedure.
You need to change them to mole
mole = mass / Molar mass
mole of Ca = 0.910 g / 40.08 g/mol = 0.0227 mol
mole of N = 0.636 g / 14.01 g/mol = 0.04539 mol
mole of O = 1.453 g / 16.00 g/mol = 0.0908 mol
Then, take the smallest number and divide for all number (in mol that I just found)
Ca = 0.0227 / 0.0227 = 1
N = 0.04539 / 0.0227 = 1.9999 (round it 2)
O = 0.0908 / 0.0227 = 4
So the empirical formula will be Ca1N2O4 ( CaN2O4 )
The mass of silver metal produced from 6.35g of copper is 21.574 grams
calculation
Cu + 2AgNO3 → Cu(NO3)2 +2 Ag
find the moles of Cu used = mass/molar mass of Cu
moles = 6.35/63.5 =0.1 Moles
by use of mole ratio between Cu to Ag which is 1:2 the moles of Ag is therefore= 0.1 x2= 0.2 moles
mass of Ag = moles x molar mass
= 0.2 moles x 107.87 g/mol = 21.574 grams
<u>Answer:</u> The partial pressure of oxygen gas is 57.8 kPa
<u>Explanation:</u>
Dalton's law of partial pressure states that the total pressure of the system is equal to the sum of partial pressure of each component present in it.
To calculate the partial pressure of oxygen gas, we use the law given by Dalton, which is:
![P_T=p_{He}+p_{O_2}+p_{CO_2}](https://tex.z-dn.net/?f=P_T%3Dp_%7BHe%7D%2Bp_%7BO_2%7D%2Bp_%7BCO_2%7D)
We are given:
Total pressure of the tank,
= 201.4 kPa
Vapor pressure of helium,
= 125.4 kPa
Vapor pressure of carbon dioxide,
= 18.2 kPa
Putting values in above equation, we get:
![201.4=125.4+p_{O_2}+18.2\\\\p_{O_2}=201.4-[125.4+18.2]=57.8kPa](https://tex.z-dn.net/?f=201.4%3D125.4%2Bp_%7BO_2%7D%2B18.2%5C%5C%5C%5Cp_%7BO_2%7D%3D201.4-%5B125.4%2B18.2%5D%3D57.8kPa)
Hence, the partial pressure of oxygen gas is 57.8 kPa