948 or 9.48 x 10^2
There are two sets of rules for significant figures
• One set for addition and subtraction
• Another set for multiplication and division
You used the set for multiplication and division.
This problem involves addition and subtraction, and the rule is
The number of places after the decimal point in the answer must be <em>no greater than the number of decimal places in every term</em> in the sum.
Thus, we have
78.9
+890.43
-21.
= 948.33
The "21" term has the fewest digits after the decimal point (none), so the answer must have no digits after the decimal point.
To the correct answer is 948 = 9.48 x 10^2. It has three significant figures.
The balanced equation for the above reaction is
2Al + 3CuCl₂ --> 2AlCl₃ + 3Cu
stoichiometry of Al to CuCl₂ is 2:3
limiting reactant is when the reactant is fully consumed in the reaction therefore amount of product formed depends on amount of limiting reactant present.
number of Al moles - 0.5 g / 27 g/mol = 0.019 mol
number of CuCl₂ moles - 3.5 g / 134.5 g/mol = 0.026 mol
if Al is the limiting reactant
if 2 mol of Al reacts with 3 mol of CuCl₂
then 0.019 mol of Al reacts with - 3/2 x 0.019 = 0.029 mol of CuCl₂
but only 0.026 mol of CuCl₂ is present
therefore CuCl₂ is the limiting reactant
and 0.026 mol of CuCl₂ reacts with - 0.026/3 x 2 = 0.017 mol of Al is required
but 0.019 mol of Al is present
therefore CuCl₂ is the limiting reactant and Al is in excess
Answer:
0.55 mol Au₂S₃
Explanation:
Normally, we would need a balanced equation with masses, moles, and molar masses, but we can get by with a partial equation, if the S atoms are balanced.
1. Gather all the information in one place:
M_r: 34.08
Au₂S₃ + … ⟶ 3H₂S + …
m/g: 56
2. Calculate the moles of H₂S
Moles of H₂S = 56 g H₂S × (34.08 g H₂S/1 mol H₂S)
= 1.64 mol H₂S
3. Calculate the moles of Au₂S₃
The molar ratio is 1 mol Au₂S₃/3 mol H₂S.
Moles of Au₂S₃ = 1.64 mol H₂S × (1 mol Au₂S₃/3 mol H₂S)
= 0.55 mol Au₂S₃
Answer:
the answer is unsaturated
Explanation:
A saturated solution contains more solute per volume of solvent than an unsaturated solution. The solute has dissolved until no more can, leaving undissolved matter in the solution. ... In a supersaturated solution, there is more dissolved solute than in a saturated solution.
Answer:
The amount of NaOH required to prepare a solution of 2.5N NaOH.
The molecular mass of NaOH is 40.0g/mol.
Explanation:
Since,
NaOH has only one replaceable -OH group.
So, its acidity is one.
Hence,
The molecular mass of NaOH =its equivalent mass
Normality formula can be written as:
Substitute the given values in this formula to get the mass of NaOH required.
Hence, the mass of NaOH required to prepare 2.5N and 1L. solution is 100g
