86 percent is the percent yield for this experiment if he expected to produce 5g of product.
Explanation:
Given that:
mass of test tube = 5 grams
mass of test tube + reactant is 12.5 grams
mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)
mass of reactant = 12.5 -5
= 7.5 grams
when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.
so mass of product formed = 9.3 - 5
= 4. 3 grams of product is formed (actual yield)
However, he expected the product to be 5 grams (theoretical yield)
Percent yield = 
x 100
putting the values in the formula:
percent yield = 
x 100
= 86 %
86 percent is the percent yield.
Answer:
Amount of HCL = 0.00318 L of 3.18 ml
Explanation:
Given:
HCL = 2.5 M
NaOH = 0.53 M
Amount of NaOH = 15 ml = 0.015 L
Find:
Amount of HCL
Computation:
HCL react with NaOH
HCl + NaOH ⇒ NaCl + H₂O
So,
Number of moles = Molarity × volume
Number of moles of NaOH = 0.53 × 0.015
Number of moles of NaOH = 0.00795 moles
So,
Number of moles of HCl needed = 0.00795 mol
es
So,
Volume = No. of moles / Molarity
Amount of HCL = 0.00795 / 2.5
Amount of HCL = 0.00318 L of 3.18 ml
First write the balanced equation of this reaction:
2H2 + O2 —> 2H2O
mol of H2= 0.60 gH2/2.02 gH2 = 0.297 mol
There are 2 mol of H2 for every 2 mol of H2O so the number of mol of H2 is equal to the number of mol of H2O.
g of H2O = 0.297 mol H2O • 18.02 gH2O = 5.35 g H2O
Do the same thing for O2:
mol of O2 = 4.8 gO2/32.0 gO2 = 0.15 mol of O2
There is 1 mol of O2 for every 2 mol of H2O so multiply 0.15 • 2 to get the number of mol of H2O
g of H2O = 0.30 mol H2O • 18.02 gH2O = 5.41 g H2O
The correct answer is 5.35 g H2O (or 5.4 g if checking significant figures) because O2, in this case, is the limiting reactant of this reaction.
The question is incomplete, the complete question is shown in the image attached to this answer.
Answer:
139.13
Explanation:
The average atomic mass of the element Likhitium is the sum of the relative abundance of all the isotopes of Likhitium.
We obtain the relative atomic mass of Likhitium as follows;
(44.7/100 * 138) + (52.3/100 * 139) + (0.5/100 * 140) + (2.5/100 * 141)
61.7 + 73.2 + 0.7 + 3.53 = 139.13
Hence the relative abundance of Likhitium is 139.13
Answer:
Analog signals are used to convert sound waves into electronic signals.
Explanation:
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