Hi, P.S. The atomic number will always be the same as the number of protons...
Answer:
The volume of the sample is 17.4L
Explanation:
The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:
0.1800mol + 0.1800mol reactants =
0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.
Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:
V1n2 = V2n1
<em>Where V is volume and n moles of 1, initial state and 2, final state of the gas</em>
Replacing:
V1 = 23.2L
n2 = 0.2700 moles
V2 = ??
n1 = 0.3600 moles
23.2L*0.2700mol = V2*0.3600moles
17.4L = V2
<h3>The volume of the sample is 17.4L</h3>
The limiting reactant when 5.6 moles of aluminium react with 6.2 moles of water is
water( H2O)
<u><em>Explanation</em></u>
The balanced equation is as below
2 Al +3 H2O → Al2O3 +3 H2
The mole ratio of Al :Al2O3 is 2:1 therefore the moles of Al2O3
= 5.6 x1/2 = 2.8 moles
The mole ratio of H2O: Al2O3 is 3:1 therefore the moles of Al2O3 produced
= 6.2 x1/3= 2.067 moles
since H2O yield less amount of Al2O3 , H2O is the limiting reagent.
Answer:
As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.
Explanation:
Absorbance of light by a reagent of concentration c, is given as
A = εcl
A = Absorbance
ε = molar absorptivity
c = concentration of reagent.
l = length of light path or length of the solution the light passes through.
So, if all.other factors are held constant, If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the absorbance in a 1-cm cell.
But the reagent blank solution is called a blank solution because it lacks the given reagent. A blank solution does not contain detectable amounts of the reagent under consideration. That is, the concentration of reagent in the blank solution is 0.
Hence, the Absorbance is subsequently 0. And increasing or decreasing the path length of light will not change anything. As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.
Hope this Helps!!!
Answer:
Option B = 60,600 mg (correct option)
Explanation:
First of all we will have an idea which numbers are consider as significant.
1 = All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.
2= Leading zeros are not consider as a significant figures. e.g. 0.02 in this number only one significant figure present which is 2.
3= Zero between the non zero digits are consider significant like 105 consist of three significant figures.
4= The zeros at the right side e.g 3400 are also significant. There are four significant figures are present.
In given options, Option A 60.6 mg have 3 significant figures.
Option B have 5 significant figures.
Option C have 4 significant figures.
Option D have 3 significant figures.
Thus option b is correct option which have more significant figures.
