<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.
Answer:
Option D. KBr < KCl < NaCl
Explanation:
We'll begin by calculating the number of mole of each sample.
This can be obtained as follow:
For NaCl:
Mass = 1 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mole of NaCl =?
Mole = mass /Molar mass
Mole of NaCl = 1/58.5
Mole of NaCl = 0.0171 mole
For Kbr:
Mass = 1 g
Molar mass of KBr = 39 + 80 = 119 g/mol
Mole of KBr =?
Mole = mass /Molar mass
Mole of KBr = 1/119
Mole of KBr = 0.0084 mole
For KCl:
Mass = 1 g
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Mole of KCl =?
Mole = mass /Molar mass
Mole of KCl = 1/74.5
Mole of KCl = 0.0134 mole
Summary
Sample >>>>>>>> Number of mole
NaCl >>>>>>>>>> 0.0171
KBr >>>>>>>>>>> 0.0084
KCl >>>>>>>>>>> 0.0134
Arranging the number of mole of the sampl in increasing order, we have:
KBr < KCl < NaCl
Answer:
CH4 + 2 O2 - - - > CO2 + 2 H2O
Explanation:
hope this helps,
you can check by adding all the numbers on both sides
please mark it
Answer:
C. 4.00 K
Explanation:
We can solve this using Charles's Law of the ideal gas. The law describes that when the pressure is constant, the volume will be directly proportional to the temperature. Note that the temperature here should only use the Kelvin unit. Before compressed, the volume of the gas is 50ml(V1) and the temperature is 20K (T1). After compressed the volume becomes 10ml(V2). The calculation will be:
V1 / T1= V2 / T2
50ml / 20K = 10ml / T2
T2= 10ml/ 50ml * 20K
T2= 4K
Answer:
0.006342moles
Explanation:
1000ml of NaOH contain 0.151moles
42ml of NaOH contain (42*0.151)/1000 moles
=0.006342moles
