Answer:
Number of moles = 10.6 mol
Explanation:
Given data:
Molar mass of H = 1.008 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Mass of citric acid = 2.03 kg (2.03×1000 = 2030 g)
Number of moles of citric acid = ?
Solution:
Formula:
Number of moles = mass/molar mass
Now we will calculate the molar mass of citric acid:
C₆H₈O₇ = (12.01× 6) + (1.008×8) + (16.00×7)
C₆H₈O₇ = 72.06 + 8.064+112
C₆H₈O₇ = 192.124g/mol
Number of moles = 2030 g/ 192.124g/mol
Number of moles = 10.6 mol
Answer:
The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are
Ku = 38.252 W/mK
K lower = 0.199 W/mK
Explanation:
As we know
Ku = Vp * Kair + Vmagnesium * K metal
Ku = 0.10 *0.02 + (1-0.25) * 51
Ku = 38.252 W/mK
The lower limit
K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)
K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)
K lower = 0.199 W/mK
Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole
Moles = 0.246
Volume = 280 mL = 0.280 L
Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g
Moles = 1.98
Volume = 2.6 L
Molarity = 0.762 M
Molarity = 0.76 M
