An unopened 350 mL can of diet Dr. Pepper will float in a lake. However, it’s also true that an unopened 350 mL can f regular Dr
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Answer:
Use a ratio of 0.44 mol lactate to 1 mol of lactic acid
Explanation:
John could prepare a lactate buffer.
He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.
He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.
For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.
The question is incomplete. There's missing the image, which is shown below.
Answer:
Volume of O₂ = 6 L, volume of mixture: 18 L, volume of H₂O = 12 L, molecule volume of H₂O = 0.667 molecule/L
Explanation:
The reaction between hydrogen gas and oxygen gas to form water is:
2H₂(g) + O₂(g) → 2H₂O(g)
So, for 1 mol of O₂ is necessary 2 moles of H₂ form 2 moles of H₂O. As the images below there's 8 molecules of H₂, 4 molecules of O₂, 12 molecules in the mixture, and 8 molecules of H₂O. Thus, there are stoichiometric values.
All the images are at the same temperature and pressure, so, by the ideal gas law:
PV= nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
The number of moles and molecules are related, so let's substitute it in the equation. For the H₂:
P*12 = 8*RT
RT/P = 12/8 = 1.5
Thus, for O₂:
PV= nRT
V = n*(RT/P)
V = 4*1.5 = 6 L
For the mixture:
V = 12*1.5 = 18 L
For H₂O:
V = 8*1.5 = 12 L
The molecule volume is the number of molecules divided by the volume they occupy, thus for water: 8/12 = 0.667 molecules/L
