Answer:
1.26x10^25 atoms of hydrogen
Explanation:
because there are 12 atoms of hydrogen in a molecule of glucose, multiply 12 by Avogadro's number (6.02x10^23) to get how many molecules of hydrogen there are in a mole of glucose. Then multiply that number by 1.75, which is the number of moles of glucose there is in this problem.
Answer:
Option a.
0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i
Explanation:
To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)
These are the colligative properties:
ΔP = P° . Xm . i → Lowering vapor pressure
ΔT = Kb . m . i → Boiling point elevation
ΔT = Kf . m . i → Freezing point depression
π = M . R . T → Osmotic pressure
Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.
CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:
CaCl₂ → Ca²⁺ + 2Cl⁻
We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3
KNO₃ → K⁺ + NO₃⁻
We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2
Option a, is the best.
Answer:
pH change is -0.07
Explanation:
Using H-H equation for acetic acid:
pH = pKa + log [Acetate salt] / [Acetic acid]
Replacing:
pH = 4.74 + log[1.188M] / [1.188M]
pH = 4.74
The HCl reacts with sodium acetate producing acetic acid, thus:
HCl + CH₃COONa → CH₃COOH + NaCl
That means the final moles of sodium acetate are initial moles - moles of HCl and moles of acetic acid are initial moles + moles of HCl.
As the volume of the buffer is 1.0L, initial moles of both substances are 1.188moles. After reaction, the moles are:
sodium acetate: 1.188mol - 0.1mol = 1.088mol
Acetic acid: 1.188mol + 0.1mol = 1.288mol
Using again H-H equation:
pH = 4.74 + log[1.088M] / [1.288M]
pH = 4.67
pH change is: 4.67 - 4.74 = -0.07
The correct answer is A. The image shows a nuclear fission. This takes place in any of the heavy nuclei after capture of a neutron. This is the opposite of nuclear fusion. In this case, nuclei are broken down into two.
A gas with a vapor density greater than that of air, would be most effectively displaced out off a vessel by ventilation.
The two following principles determine the type of ventilation: Considering the impact of the contaminant's vapour density and either positive or negative pressure is applied.
Consider a vertical tank that is filled with methane gas. Methane would leak out if we opened the top hatch since its vapour density is far lower than that of air. A second opening could be built at the bottom to greatly increase the process' efficiency.
A faster atmospheric turnover would follow from air being pulled in via the bottom while the methane was vented out the top. The rate of natural ventilation will increase with the difference in vapour density. Numerous gases that require ventilation are either present in fairly low concentrations or have vapor densities close to one.
