Explanation:
As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.
So, using ideal gas equation we will calculate the number of moles as follows.
PV = nRT
or, n = ![\frac{PV}{RT}](https://tex.z-dn.net/?f=%5Cfrac%7BPV%7D%7BRT%7D)
= ![\frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B756%7D%7B760%7Datm%20%5Ctimes%200.129%20L%7D%7B0.0821%20Latm%2Fmol%20K%20%5Ctimes%20298%20K%7D)
= 0.0052 mol
Also, No. of moles = ![\frac{mass}{\text{molar mass}}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7B%5Ctext%7Bmolar%20mass%7D%7D)
0.0052 mol = ![\frac{mass}{2 g/mol}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7B2%20g%2Fmol%7D)
mass = 0.0104 g
As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at
= 24 mm Hg
=
atm
= 0.03158 atm
Now, P = ![\frac{756}{760} - 0.03158](https://tex.z-dn.net/?f=%5Cfrac%7B756%7D%7B760%7D%20-%200.03158)
= 0.963 atm
Hence, n =
= 0.0056 mol
So, mass of
= 0.0056 mol × 2
= 0.01013 g (actual yield)
Therefore, calculate the percentage yield as follows.
Percent yield = ![\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BActual%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%20%5Ctimes%20100)
=
= 97.49%
Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.
Answer:
H₂O + CO₂ → H₂CO₃
Option D is correct.
Law of conservation of mass:
According to this law, mass can neither be created nor destroyed in a chemical equation.
This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Now we will apply this law to given chemical equations:
A) H₂ + O₂ → H₂O
There are two H and two O atoms present on left side while on right side only one O and two H atoms are present so mass in not conserved. This option is incorrect.
B) Mg + HCl → H₂ + MgCl₂
In this equation one Mg, one H and one Cl atoms are present on left side of equation while on right side two H, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.
C) KClO₃ → KCl + O₂
There are one K, one Cl and three O atoms are present on left side of equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.
D) H₂O + CO₂ → H₂CO₃
There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. This option is correct.
Answer:
P = 13.5 atm
Explanation:
Given that
No. of moles, n = 20 moles
Volume of nitrogen gas = 36.2 L
Temperature = 25°C = 298 K
We need to find the pressure of the gas. Using the ideal gas equation
PV = nRT
Where
R is gas constant, ![R=0.082057\ L-atm/K-mol](https://tex.z-dn.net/?f=R%3D0.082057%5C%20L-atm%2FK-mol)
So,
![P=\dfrac{nRT}{V}\\\\P=\dfrac{20\times 0.082057\times 298}{36.2 }\\\\P=13.5\ atm](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7BnRT%7D%7BV%7D%5C%5C%5C%5CP%3D%5Cdfrac%7B20%5Ctimes%200.082057%5Ctimes%20298%7D%7B36.2%20%7D%5C%5C%5C%5CP%3D13.5%5C%20atm)
so, the pressure of the gas is equal to 13.5 atm.
Answer:
A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)
Explanation:
Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.
The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:
Ca(s), C(s) and O₂(g)
That means, the equation that represents standard enthalpy of CaCO₃ is:
<h3>A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)</h3><h3 />
<em>Is the equation that has ΔH° = -1207kJ/mol</em>
Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)
The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.