The temperature : 263.016 K
<h3>Further explanation</h3>
Combined with Boyle's law and Gay Lussac's law
P1 = initial gas pressure (N/m² or Pa)
V1 = initial gas volume (m³)
P2 = gas end pressure
V2 = the final volume of gas
T1 = initial gas temperature (K)
T2 = gas end temperature
P1=760 mmHg
V1= 4 L
T1 = 275 K
P2=704 mmHg
V1=4.13 L
Answer:
The percent yield is 66.7%
Explanation:
1. The balanced chemical reaction to obtain carbon dioxide from the methane is:
2. Calculathe the theoretical quantity of carbon dioxide.
As the problem says that the limiting reagent is the methane, all the calculations will be made from this quantity:
3. Calculate the percent yield:
Percent yield = 
Percent yield = 
Percent yield = 66.7%
Explanation:
First, we need to calculate the number of moles of sodium carbonate we have in a 25 g sample. To calculate this, we will
find the molar mass of sodium carbonate (Na2CO3):
⇒ 2 × Molar mass of sodium + Molar mass of carbon + 3×molar mass of oxygen
⇒ 2 × 23 + 12 + 3 × 16
⇒ 46 + 12 + 48
⇒ 106g/mol
Thus, the molar mass of Na2CO3 is 106g/mol.
Therefore, number of moles = 25 ÷ 106
=> 0.2358 mol
Now, we know that every mole of Na2CO3 have 0.2358 moles of Na+ ions. Hence, total moles of Na2CO3 is 0.4716 moles
Number of ions present = 6.022 × 1023 × 0.4716 mol = 2.84 × 1023ions
Answer:
Standard reduction potential is an intensive property---- True
Reduction takes place at the anode ----- False
The half reaction with the lower standard reduction potential will be at the cathode in a galvanic cell ------false
The half reaction with the higher standard reduction potential will be at the cathode in a galvanic cell ------ True
Explanation:
An intensive property is a property of a substance which is inherent in it and part of its nature. It does not depend on the amount of substance present in the substance. Standard reduction potential is an intensive property.
In a galvanic cell, oxidation takes place at the anode and reduction takes place at the cathode. At the anode, the electrode potential is more negative (an oxidation) while at the cathode the reduction potential is less negative (a reduction).
Answer:
V = 15.9512 dm³
Explanation:
Given data:
Pressure = P = 1.37 atm
Temperature = T= 315 K
Number of moles of nitrogen= n = 0.845 mol
Volume = V = ?
Formula:
PV = nRT
Now we will put the values in equation.
V = nRT/ P
V = ( 0.845 mol× 0.0821 dm³.atm.K⁻¹.mol⁻¹ × 315 K) / 1.37 atm
V = 21.853 dm³. atm/ 1.37 atm
V = 15.9512 dm³
