Answer:
The answer is 130.953 g of hydrogen gas.
Explanation:
Hydrogen gas is formed by two atoms of hydrogen (H), so its molecular formula is H₂. We can calculate is molecular weight as the product of the molar mass of H (1.008 g/mol):
Molecular weight H₂= molar mass of H x 2= 1.008 g/mol x 2= 2.01568 g
Finally, we obtain the number of mol of H₂ there is in the produced gas mass (264 g) by using the molecular weight as follows:
mass= 264 g x 1 mol H₂/2.01568 g= 130.9731703 g
The final mass rounded to 3 significant digits is 130.973 g
Answer:
HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)
NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)
Explanation:
A buffer is a solution that resists changes in acidity or alkalinity. A buffer is able to neutralize a little amount of acid or base thereby maintaining the pH of the system at a steady value.
A buffer may be an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid.
The equations for the neutralizations that occurred upon addition of HCl or NaOH are;
HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)
NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)
Answer:
Explanation:
Arrhenius defined an acid as a substance that interacts with water to produce excess hydrogen ions in aqueous solution.
A base is a substance which interacts with water to yield excess hydroxide ions, in an aqueous solution according to Arrhenius.
Bronsted-Lowry theory defined an acid as a proton donor while a base is a proton acceptor.
Answer:
LiOH+CH3COOH=C2H3LiO2+H2O balanced equation
Explanation:
lithium hydroxide-> LiOH
hydrogen acetate->CH3COOH
lithium acetate->C2H3LiO2
water->H2O
LiOH+CH3COOH=C2H3LiO2+H2O
Li=1,O=3,H=5,C=2 --> Li=1,O=3,H=5,C=2
the two sides have same numbers so it is balanced.
Answer:
25.8 L
Explanation:
Step 1: Write the balanced equation for the complete combustion of acetylene
C₂H₂ + 2.5 O₂ ⇒ 2 CO₂ + H₂O
Step 2: Calculate the moles corresponding to 15.0 g of C₂H₂
The molar mass of C₂H₂ is 26.04 g/mol.
15.0 g × 1 mol/26.04 g = 0.576 mol
Step 3: Calculate the moles of CO₂ produced from 0.576 moles of C₂H₂
The molar ratio of C₂H₂ to CO₂ is 1:2. The moles of CO₂ produced are 2/1 × 0.576 mol = 1.15 mol.
Step 4: Calculate the volume corresponding to 1.15 moles of CO₂
At standard temperature and pressure, 1 mole of CO₂ occupies 22.4 L.
1.15 mol × 22.4 L/1 mol = 25.8 L
