For the equation to balance out, you would need 2 H2O on the product. That will give a total of 1 Carbon, 4 Hydrogens and 4 Oxygens on each side.
Answer:
[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)
Explanation:
1. Molarity = moles solute / Volume solution in Liters
=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH
=> volume of solution (assuming density of final solution is 1.0g/ml) ...
volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution
Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH
2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)
From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln
= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.
The balanced reaction is:
N2 + 3H2 = 2NH3
We are given the amount of the product to be produced.This will be the starting point of our calculations. We use the ideal gas equation to find for the number of moles.
<span>
n = PV / RT = 1.00(.520 L) / (0.08206 atm L/mol K ) 273 K
n= 0.0232 mol NH3
</span>0.0232 mol NH3 (1 mol N2 / 2 mol NH3) = 0.0116 mol N2
<span>Therefore, the correct answer is A.</span>
The molar extinction coefficient is 15,200
.
The formula to be used to calculate molar extinction coefficient is -
A = ξcl, where A represents absorption, ξ refers molar extinction coefficient, c refers to concentration and l represents length.
The given values are in required units, hence, there is no need to convert them. Directly keeping the values in formula to find the value of molar extinction coefficient.
Rewriting the formula as per molar extinction coefficient -
ξ = 
ξ = 
Performing multiplication in denominator to find the value of molar extinction coefficient
ξ =
Performing division to find the value of molar extinction coefficient
ξ = 15,200 
Hence, the molar extinction coefficient is 15,200
.
Learn more about molar extinction coefficient -
brainly.com/question/14744039
#SPJ4
Answer:
the ph of an aqueous solution of sulphuric acid which is 5*10^5 mol in concentration is basic in nature