Answer:
For these problems, we need to compare the theoretical yield that we'd get from performing stoichiometry to the actual yield stated in the problem. % yield is the actual yield/theoretical yield x 100%
Cu + 2 AgNO₠→ Cu(NOâ‚)â‚‚ + 2 Ag ==> each mole of copper yields two moles of silver
12.7-g Cu x ( 1 mol Cu /63.5-g Cu) x ( 2 mol Ag / 1 mol Cu) x (108-g Ag / 1 mol Ag) = 43.2-g Ag. This is the theoretical yield. Now, since we got 38.1-g Ag our % yield is:
38.1-g/43.2-g x 100% = 88.2%
Answer:
hello your question lacks the required reaction pairs below are the missing pairs
Reaction system 1 :
A + B ⇒ D ![-r_{1A} = 10exp[-8000K/T]C_{A}C_{B}](https://tex.z-dn.net/?f=-r_%7B1A%7D%20%20%3D%2010exp%5B-8000K%2FT%5DC_%7BA%7DC_%7BB%7D)
A + B ⇒ U 
Reaction system 2
A + B ⇒ D 
B + D ⇒ U 
Answer : reaction 1 : description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1
condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2
reaction 2 :
description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1
condition to maximize selectivity:
to increase selectivity the concentration of D should be minimal
Explanation:
Reaction system 1 :
A + B ⇒ D
A + B ⇒ U
the selectivity of D is represented using the relationship below
hence SDu = 1/10 * 
description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1
condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2
Reaction system 2
A + B ⇒ D
B + D ⇒ U
selectivity of D
hence Sdu = 
description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1
condition to maximize selectivity:
to increase selectivity the concentration of D should be minimal
Answer :
- Boiling point of the sugar solution will be higher than that of water's boling point.
- Freezing point of the sugar solution will be lower than that of water's freezing point.
Explanation:
- Boiling point of a liquid is defined as temperature at which vapor pressure of liquid becomes equal to the atmospheric pressure.
Boiling point of solution is always higher than that of the pure solvent
Vapor pressure increases with increase in temperature which means sugar solution will be heated more to make vapor pressure equal to atmospheric pressure.
- Freezing point is defined as temperature at which solid and liquid phase are at equilibrium or temperature at which vapor pressure of liquid becomes equal to the vapor pressure in its solid phase.
Freezing point of solution is always lower than that of the pure solvent.
Lower the temperature, lower will be the vapor pressure which sugar solution solution will get freeze at lower temperature than that of the water.
Answer:
Mass = 141.6 g
Explanation:
Given data:
Mass of Kr in gram = ?
Volume in L = 9.59 L
Temperature = 46.0°C
Pressure = 4.62 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
46.0+273 = 319 K
4.62 atm × 9.59 L = n× 0.0821 atm.L/ mol.K ×319 K
44.3 atm.L = n×26.19 atm.L/ mol
n = 44.3 atm.L / 26.19 atm.L/ mol
n = 1.69 mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 1.69 mol × 83.79 g/mol
Mass = 141.6 g
They are eaten by Tertiary consumers
