Answer: Mass of 
produced in this reaction was 6.56 grams
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Mass or reactants = Mass of 
+ mass of 
= 16.00 + 64.80 = 80.80 g
Mass of products = mass of aqueous solution + mass of + = 74.24 + x g
Mass or reactants = Mass of products
80.80 g = 74.24 + x g
x = 6.56 g
Thus mass of produced in this reaction was 6.56 grams
We know that there are 100 cm in 1 m, so we can use this to convert to meters:
Therefore we know that 
cm is equal to 2.41 m.
Answer:
11.66 L.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If P and T are constant, and have different values of n and V:
<em>(V₁n₂) = (V₂n₁).</em>
V₁ = 25.5 L, n₁ = 3.5 mol.
V₂ = ??? L, n₂ = 3.5 mol - 1.9 mol = 1.6 mol.
<em>∴ V₂ = (V₁n₂)/(n₁)</em> = (25.5 L)(1.6 mol)/(3.5 mol) =<em> 11.66 L.</em>
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
Answer:
He provided a number of scientific insights that laid the foundation for future scientists. He also improved telescope that helped further the understanding of the world and universe.
Explanation:
