Answer:
The resulting pressure is 2.81 atm
Explanation:
According to Dalton's Law of Partial Pressure, each of the gases (A and B) will exert their pressure independently. If we use Boyle's Law to calculate the pressure of each of the gases separately we have:
Pressure of gas A:
p1V1 = p2V2
p1 = 2.4 atm
V1 = 722 mL
V2 = 722 + 169 = 891 mL
p2 =?
Clearing p2:
p2 = (p1V1)/V2 = (2.4*722)/891 = 1.94 atm
Pressure of gas B:
p1 = 4.6 atm
V1 = 169 mL
V2 = 169+722 = 891 mL
p2=?
Clearing p:
p2 = (4.6*169)/891 = 0.87 atm
Dalton's expression for total partial pressures is equal to:
ptotal = pA + pB = 1.94+0.87 = 2.81 atm
Answer
CH3Cl is polar amoungst the choices
Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
Answer:
yes
Explanation:
oak trees don't have vertebrates
The term sensitivity in Analytical Chemistry is "the slope of the calibration curve or a function of analyte concentration or amount".
<u>Answer:</u> Option B
<u>Explanation:</u>
In a sample, the little amounts of substances can be accurately evaluated by a method is termed as "Analytical sensitivity". This detect a target analyte like an antibody or antigen, process is considered as potential of a test to and generally demonstrated as the analyte's minimum detectable concentration.
The acceptable diagnostic sensitivity is not guaranteed by high analytical sensitivity. The percentage of individuals who have a given disarray who are identified by the method as positive for the disarray is known as "Diagnostic sensitivity".
