Answer:
14.2L at STP
Explanation:
Based on the problem, 2 moles of NH3 produce 6 moles of HF. To solve this question we have to convert the mass of NH3 to moles. With the chemical equation find the moles of HF and using PV = nRT find the liters of HF:
<em>Moles NH3 -Molar mass: 17.031g/mol-</em>
3.6g NH3 * (1mol / 17.031g) = 0.211 moles NH3
<em>Moles HF:</em>
0.211 moles NH3 * (6mol HF / 2mol NH3) = 0.634 moles HF
<em>Volume HF</em>
PV = nRT; V = nRT/P
<em>Where V is volume in liters, n are moles of the gas = 0.634 moles, R is gas constant = 0.082atmL/molK, T is absolute temperature = 273.15K at STP and P is pressure = 1atm at STP.</em>
Replacing:
V = 0.634moles*0.082atmL/molK*273.15K / 1atm
V = 14.2L at STP
Answer:
The easiest way is to titrate a sample with a solution of a base.
Explanation:
Molarity of a solution if 124.86 g of rbf are dissolved into a solution of water that has a final volume of 2.00L is 0.59.
<h3>What is molarity?</h3>
Molarity is used for dilute aqueous solutions held at a constant temperature. In general, the difference between molarity and molality for aqueous solutions near room temperature is very small and it won't really matter whether you use a molar or molal concentration.
MOLARITY = no of moles of solute/volume of soln in litres
No of moles of rbf = 124.6/104.46
= 1.19
Volume of soln = 2
Molarity=1.19/2 = 0.59
Learn more about Molarity here: brainly.com/question/26756988
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Answer:
not sure if this us right but oxygen and fluorine? I learned chem but I forgot about it kinda sorry
Answer:
The mass of the precipitate that AgCl is 3.5803 g.
Explanation:
a) To calculate the molarity of solution, we use the equation:
We are given:
Mass of solute (NaCl) = 1.46 g
Molar mass of sulfuric acid = 58.5 g/mol
Volume of solution = 
Putting values in above equation, we get:
0.09982 M is the concentration of the sodium chloride solution.
b) 
Moles of NaCl = 
according to reaction 1 mol of NaCl gives 1 mol of AgCl.
Then 0.02495 moles of NaCl will give:
of AgCl
Mass of 0.02495 moles of AgCl:
The mass of the precipitate that AgCl is 3.5803 g.
