The temperature is the same but the heat flow is the opposite.
Answer : The molar mass of an acid is 266.985 g/mole
Explanation : Given,
Mass of an acid (HX) = 4.7 g
Volume of NaOH = 32.6 ml = 0.0326 L
Molarity of NaOH = 0.54 M = 0.54 mole/L
First we have to calculate the moles of NaOH.
Now we have to calculate the moles of an acid.
In the titration, the moles of an acid will be equal to the moles of NaOH.
Moles of an acid = Moles of NaOH = 0.017604 mole
Now we have to calculate the molar mass of and acid.
Now put all the given values in this formula, we get:
Therefore, the molar mass of an acid is 266.985 g/mole
Answer:
See explanation below
Explanation:
In order to calculate this, we need to use the following expression to get the concentration of the base:
MaVa = MbVb (1)
We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:
Atomic weights of the elements to be used:
K = 39.0983 g/mol; H = 1.0078 g/mol; C = 12.0107 g/mol; O = 15.999 g/mol
MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol
Now, let's calculate the mole of KHP:
moles = 0.5053 / 204.2189 = 0.00247 moles
With the moles, we also know that:
n = M*V (2)
Replacing in (1):
n = MbVb
Now, solving for Mb:
Mb = n/Vb (3)
Finally, replacing the data:
Mb = 0.00247 / (13.4473/1000)
Mb = 0.184 M
This would be the concentration of NaOH
Answer:Videos
For example, when oxygen and hydrogen react to produce water, one mole of oxygen ... These conversion factors state the ratio of reactants that react but do not tell ... In a typical chemical equation, an arrow separates the reactants on the left ... For example, to determine the number of mol
Answer:
A jump occurs when a core electron is removed.
Explanation:
A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.
For Beryllium, the electronic configuration of is 1s2 2s2.
There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron
The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron
The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron
The electronic configuration of Lithium is 1s2 2s1
There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.
