During a chemical reaction, the mass of the products is always equal to the mass of the reactants.
There are 453.592 grams in a pound.
Answer:
23.0733 L
Explanation:
The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:
![Mass=\frac {50}{100}\times 125\ g](https://tex.z-dn.net/?f=Mass%3D%5Cfrac%20%7B50%7D%7B100%7D%5Ctimes%20125%5C%20g)
Mass = 62.5 g
Molar mass of
= 34 g/mol
The formula for the calculation of moles is shown below:
Thus, moles are:
![moles= 1.8382\ mol](https://tex.z-dn.net/?f=moles%3D%201.8382%5C%20mol)
Consider the given reaction as:
![2H_2O_2_{(aq)}\rightarrow2H_2O_{(l)}+O_2_{(g)}](https://tex.z-dn.net/?f=2H_2O_2_%7B%28aq%29%7D%5Crightarrow2H_2O_%7B%28l%29%7D%2BO_2_%7B%28g%29%7D)
2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.
Also,
1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.
So,
1.8382 moles of hydrogen peroxide decomposes to give ![\frac {1}{2}\times 1.8382 mole of oxygen gas. Moles of oxygen gas produced = 0.9191 molGiven: Pressure = 746 torr The conversion of P(torr) to P(atm) is shown below: [tex]P(torr)=\frac {1}{760}\times P(atm)](https://tex.z-dn.net/?f=%5Cfrac%20%7B1%7D%7B2%7D%5Ctimes%201.8382%20mole%20of%20oxygen%20gas.%20%3C%2Fp%3E%3Cp%3EMoles%20of%20oxygen%20gas%20produced%20%3D%200.9191%20mol%3C%2Fp%3E%3Cp%3EGiven%3A%20%3C%2Fp%3E%3Cp%3EPressure%20%3D%20746%20torr%0A%3C%2Fp%3E%3Cp%3EThe%20conversion%20of%20P%28torr%29%20to%20P%28atm%29%20is%20shown%20below%3A%0A%3C%2Fp%3E%3Cp%3E%5Btex%5DP%28torr%29%3D%5Cfrac%20%7B1%7D%7B760%7D%5Ctimes%20P%28atm%29)
So,
Pressure = 746 / 760 atm = 0.9816 atm
Temperature = 27 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (27 + 273.15) K = 300.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K
<u>⇒V = 23.0733 L</u>
The volume of the gas V2 is 66.67 L
<u>Explanation:</u>
<u>Given data</u>
V1= 59.5 L
V2=?
T1= 121K Pa
T2= 108 K Pa
We have the formula,
V1/T1 =V2/T2
Rewrite the formula as,
V2= VI T1/T2
V2= 59.5 L * 121 K Pa/108 K Pa
V2=7199.5 L K Pa/ 108 K Pa
V2=66.67 L
The volume of the gas V2 is 66.67 L
5000 units of antigens are present in 0.5 mL of the mumps virus vaccine.
Volume of the vaccine after dilution with water = 2 mL
0.5 mL has 5000 units antigen. So when 0.5 mL was diluted to 2 mL, the final vaccine after dilution would contain 5000 units of antigen.
Units of antigen present in 1 mL vaccine sample
= ![1 mL *\frac{5000 units}{2 mL} = 2500 units](https://tex.z-dn.net/?f=%201%20mL%20%2A%5Cfrac%7B5000%20units%7D%7B2%20mL%7D%20%3D%202500%20units%20)
Therefore, 2500 units of the antigen will be present in each milliliter, if 0.5 mL of the vaccine is diluted to 2 mL with water for injection.