Answer:
43.2 moles of carbon dioxide are required and 421g of glucose could be produced
Explanation:
Based on the reaction:
6CO2 + 6H2O → C6H12O6 + 6O2
1 mole of glucose, C6H12O6, requires 6 moles of carbon dioxide. 7.2moles of glucose requires:
7.2mol C6H12O6 * (6mol CO2 / 1mol C6H12O6) =
<h3>43.2 moles of carbon dioxide are required</h3><h3 />
618g of CO2 -Molar mass: 44.01g/mol- are:
618g * (1mol / 44.01g) = 14.04moles CO2
Moles C6H12O6:
14.04moles CO2 * (1mol C6H12O6 / 6mol CO2) = 2.34moles C6H12O6
Mass glucose -Molar mass: 180.156g/mol-
2.34moles C6H12O6 * (180.156g / mol) =
<h3>421g of glucose could be produced</h3>
Co2 will dissolve in water if water is saturated with Co2 first then it will absorb some of CO2 it appear as less is absorb that was actually the case . this would make it appear as present of CACO3 is in original sample is lower than it will be too low
Answer:
Test the pH of things like coffee, spit, and soap to determine whether each is acidic, basic, or neutral. Visualize ... Investigate whether changing the volume or diluting with water affects the pH. ... Original Sim and Translations ... Lab, Chemistry.
Answer:
n = 2.58 mol
Explanation:
Given data:
Number of moles of argon = ?
Volume occupy = 58 L
Temperature = 273.15 K
Pressure = 1 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
1 atm × 58 L = n × 0.0821 atm.L/ mol.K × 273.15 K
58 atm.L = n × 22.43 atm.L/ mol.
n = 58 atm.L / 22.43 atm.L/ mol
n = 2.58 mol
