Answer:
For 0.353 moles AgNO3, we'll have 0.353 moles AgCl
Explanation:
How many moles of AgCl will be produced from 60.0g AgNO3 assuming NaCl is available in excess.
Step 1: Data given
Mass of AgNO3 = 60.0 grams
Molar mass AgNO3 = 169.87 g/mol
NaCl is in excess, so AgNO3 is the limiting reactant
Step 2: The balanced equation
AgNO3 + NaCl → AgCl + NaNO3
Step 3: Calculate moles AgNO3
Moles AgNO3 = mass AgNO3 / molar mass AgNO3
Moles AgNO3 = 60.0 grams / 169.87 g/mol
Moles AgNO3 = 0.353 moles
Step 4: Calculate moles AgCl
For 1 mol AgNO3 we need 1 mol NaCl to produce 1 mol AgCl and 1 mol NaNO3
For 0.353 moles AgNO3, we'll have 0.353 moles AgCl
Answer:
The correct answer is - 5 carbon compounds due to low to high intermolecular forces between their molecules.
Explanation:
Bottle C has gas in it and we know that alkane has carbon and hydrogen only which means they have a single sigma bond between them and very low intermolecular forces in between molecules and are present mostly at gaseous state. Thus, bottle C has alkane.
Alcohols have -OH group that can form rarely two pi bonds which means they have intermediate intermolecular force whereas acids have -cooH group with a high molecular force so bottle B with liquid is alcohol and A has acid.
Answer: B. CH₂OH
Explanation:
First off, we need a liquid solvent. Eliminate C and D.
Now, we have to consider polarity.
- Carbon tetrachloride is nonpolar, so sodium chloride will be insoluble.
- CH₂OH is polar, so sodium chloride will be soluble.
The answer is d.HCl hope this helps
Answer:
a) Keq = 4.5x10^-6
b) [oxaloacetate] = 9x10^-9 M
c) 23 oxaloacetate molecules
Explanation:
a) In the standard state we have to:
ΔGo = -R*T*ln(Keq) (eq.1)
ΔGo = 30.5 kJ/moles = 30500 J/moles
R = 8.314 J*K^-1*moles^-1
Clearing Keq:
Keq = e^(ΔGo/-R*T) = e^(30500/(-8.314*298)) = 4.5x10^-6
b) Keq = ([oxaloacetate]*[NADH])/([L-malate]*[NAD+])
4.5x10^-6 = ([oxaloacetate]/(0.20*10)
Clearing [oxaloacetate]:
[oxaloacetate] = 9x10^-9 M
c) the radius of the mitochondria is equal to:
r = 10^-5 dm
The volume of the mitochondria is:
V = (4/3)*pi*r^3 = (4/3)*pi*(10^-15)^3 = 4.18x10^-42 L
1 L of mitochondria contains 9x10^-9 M of oxaloacetate
Thus, 4.18x10^-42 L of mitochondria contains:
molecules of oxaloacetate = 4.18x10^-42 * 9x10^-9 * 6.023x10^23 = 2.27x10^-26 = 23 oxaloacetate molecules
