Solid, Liquid, Gas, and Plasma
4Ag + 2H2S + O2 -> 2Ag2S + 2H20 i’m pretty sure that’s correct
Answer:
The answer to your question is 8.28 g of glucose
Explanation:
Data
Glucose (C₆H₁₂O₆) = ?
Ethanol (CH₃CH₂OH)
Carbon dioxide (CO₂) = 2.25 l
Pressure = 1 atm
T = 295°K
Reaction
C₆H₁₂O₆ ⇒ 2C₂H₅OH(l) +2CO₂(g)
- Calculate the number of moles
PV = nRT
Solve for n
Substitution
Simplification
n = 0.092
- Calculate the mass of glucose
1 mol of glucose --------------- 2 moles of carbon dioxide
x --------------- 0.092 moles
x = (0.092 x 1) / 2
x = 0.046 moles of glucose
Molecular weight of glucose = 180 g
180 g of glucose --------------- 1 mol
x g ---------------0.046 moles
x = (0.046 x 180) / 1
x = 8.28 g of glucose
Answer:
both Mg isotopes have same atomic number which is equal to number of protons , and the chemical properties of an element depends upon the the number of protons in its atom . and both Mg 24 12 and Mg 26 12 have same number of protons hence they exhibits similar chemical
okay
Explanation:
i hope it helped you .
Answer:
44.8 L
Explanation:
Ideal Gas Equation -
i.e.,
PV = nRT
where,
P = pressure
V = volume
n = moles
R = universal gas constant
T = temperature
Using the information given in the question, Volume of the gas can be calculated -
P = 101.3 kPa
V = ?
n = 2.00 moles
R = 8.31
T = 0 degree C = 273.15 K
Using the above data, and putting the data in the respective formula -
PV = nRT
101.3 kPa * V = 2.00 moles * 8.31 * 273.15 K
V = 44.8 L
Hence, the volume of the given gas = 44.8 L