Molar mass ethanol:
C2H5OH = 12 x 2+ 1 x 5 + 16 + 1 = 46.0 g/mol
volume = 545 mL in liters: 545 / 1000 => 0.545 L
number of moles:
29.0 / 46.0 => 0.6304 moles
M = n / V
M = 0.6304 / 0.545
M = 1.156 mol/L
hope this helps!
Hey there!:
Given the reaction:
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
5 moles O2 ------------- 4 moles CO2
3.00 moles O2 ---------- ( moles of CO2 ?? )
moles of CO2 = 3.00 * 4 / 5
moles of CO2 = 12 / 5
moles of CO2 = 2.4 moles
So, molar mass CO2 = 44.01 g/mol
Therefore:
1 mole CO2 -------------- 44.01 g
2.4 moles CO2 ---------- ( mass of CO2 )
mass of CO2 = 2.4 * 44.01 / 1
mass of CO2 = 106 g
Answer A
Hope that helps!
To turn the flow of electricity on or off. Probably wrong
Answer:
that the poem needs to be finished heh pog
Explanation:
Answer:
False
Explanation:
The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE
This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺
