Answer:
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
Explanation:
A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.
To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:
the molar mass of water is:
H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole
So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?
moles of water= 0.1728
Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?
heat= 7.026 kJ
<u><em>The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.</em></u>
Helium gas would be the one to effuse faster as compared to krypton. This is because krypton is heavier than helium. Heavier molecules would require more energy to move so they tend to effuse slower than molecules that are lighter which will only require less energy.
Answer: we learned this not to long ago i think its a,c
Explanation:
Answer:
Single displacement and reduction
Explanation:
In a single-displacement reaction, one element exchanges partners with another.
This is a single-displacement reaction, because the element Ca exchanges partners with H.
This is also a reduction/oxidation (redox) reaction, because the optically active of Ca increases from 0 to +2 (oxidation), while the oxidation number of H decreases from +1 to 0 (reduction),
The most common types of reactions are:
- Combination
- Decomposition
- Single displacement
- Double displacement
- Reduction/oxidation
Answer:
pH = 4.71
Explanation:
We can find the pH of a buffer (Mixture of weak acid: CH3COOH, and its conjugate base: CH3COONa) using H-H equation:
pH = pKa + log [CH3COONa] / [CH3COOH]
<em>Where pH is the pH of the buffere = 4.74, pKa the pka of the buffer and [] could be taken as the moles of each reactant.</em>
As initially [CH3COONa] = [CH3COOH], [CH3COONa] / [CH3COOH] = 1:
pH = pKa + log 1
4.74 = pKa
To solve this question we need to find the initial moles of each species, The CH3COONa reacts with HCl to produce CH3COOH. That means the moles of CH3COOH after the reaction are: Initial CH3COOH + Moles HCl
Moles CH3COONa: Initial CH3COONa - Moles HCl.
<em>Moles CH3COOH: </em>
0.100L * (0.50mol / L) = 0.050 moles CH3COOH + 0.0020 moles HCl =
0.052 moles CH3COOH
<em>Moles CH3COONa: </em>
0.100L * (0.50mol / L) = 0.050 moles CH3COONa - 0.0020 moles HCl =
0.048 moles CH3COONa
Using H-H equation:
pH = 4.74 + log [0.048 moles] / [0.052 moles]
<h3>pH = 4.71</h3>
