Answer : The amount of salt present in the tank initially are, 50 kg
Explanation : Given,
Mass of salt in the tank = 50 kg
Mass of water in the tank = 1000 L
Rate of flow of water in the tank = 6 L/min
Rate of flow of solution in the tank = 3 L/min
Now we have to determine the amount of salt present in the tank initially.
As per question, there are 50 kg of salt present in the tank initially in 1000 L of water.
Thus, the amount of salt present in the tank initially are, 50 kg
the answer would be protons
Number of Atoms in Methanol are calculated as,
Write chemical formula of Methanol,
CH₃-OH
Find out Subscripts:
There is only one subscript (₃) written after Hydrogen, it means there are 3 Hydrogen atoms attached to Carbon atoms.
Add all atoms in molecule:
Carbons = 1
Hydrogens = 4
Oxygens = 1
Total Atoms ______
6 Atoms
Answer:
46 g
Explanation:
The balanced equation of the reaction between O and NO is
2 NO + O₂ ⇔ 2 NO₂
Now, you need to find the limiting reagent. Find the moles of each reactant and divide the moles by the coefficient in the equation.
NO: (80 g)/(30.006 g/mol) = 2.666 mol
(2.666 mol)/2 = 1.333
O₂: (16 g)/(31.998 g/mol) = 0.500 mol
(0.500 mol)/1 = 0.500 mol
Since O₂ is smaller, this is the limiting reagent.
The amount of NO₂ produced will depend on the limiting reagent. You need to look at the equation to determine the ratio. For every mole of O₂ reacted, 2 moles of NO₂ are produced.
To find grams of NO₂ produced, multiply moles of O₂ by the ratio of NO₂ to O₂. Then, convert moles of NO₂ to find grams.
0.500 mol O₂ × (2 mol NO₂/1 mol O₂) = 1.000 mol NO₂
1.000 mol × 46.005 g/mol = 46.005 g
You will produce 46 g of NO₂.
