Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
Water has the special type of attraction called Hydrogen bonding. The bonds between the Hydrogen and the Oxygen in each water molecule make a super dipole because the Oxygen atom is way more electronegative than the hydrogen atom. These OH bonds can then be attracted to other H2O molecules. If you have ever poured water up to the brim and there is little bit of water that is poking above the top, hydrogen bonding keeps those water molecules from spilling
Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.
The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.
When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.
Learn more: brainly.com/question/2510654
Complete Question:
A chemist adds 55.0 mL of a 1.1M barium acetate (Ba(C2H3O2)2) solution to a reaction flask. Calculate the mass in grams of barium acetate the chemist has added to the flask. Round your answer to 2 significant digits.
Answer:
15 g
Explanation:
The concentration of the barium acetate is given in mol/L (M), thus, the number of moles (n) of it is the concentrantion multiplied by the volume (55.0 mL = 0.055 L):
n = 1.1 * 0.055
n = 0.0605 mol
The molar mass of the substance can be calculated by the sum of the molar mass of each element, which can be found at the periodic table. Thus:
Ba = 137.33 g/mol
C = 12.00 g/mol
H = 1.00 g/mol
O = 16.00 g/mol
Ba(C2H3O2)2 = 137.33 + 4*12 + 6*1 + 4*16 = 255.33 g/mol
The molar mass is the mass divided by the number of moles, thus the mass (m) is the molar mass multiplied by the number of moles.
m = 255.33 * 0.0605
m = 15.45 g
Rounded by 2 significant digits, m = 15 g.
