Answer: The transition metals make up the middle block of he periodic table
The statement “Only the “Conclusion” section discusses whether the original hypothesis was supported, and both sections suggest further research”, best describes the difference between analysis and conclusion.
Answer: Option 4
<u>Explanation:
</u>
In research, we do experiments and derive the results. Then, those results were analyzed by us. In this analysis part, we compare our results with the related results published elsewhere. Also, we correlate the similarities and point out the differences between our analysis and other reported results.
In conclusion part, we have to check hypothesis or it supported. And, we summarise our analysis and figure out the further research need to be done on that to improvise our research. So, the final statement is the correct option which best describes the difference between analysis and conclusion.
Answer:
C) hydrogen bonding
Explanation:
All atoms and molecules have London Dispersion Forces between them, but they are usually overshadowed but the much stronger forces. In this scenario the major attractive force in HF molecules are hydrogen bonds. Hydrogen bonds are electrostatic forces of attraction found when Hydrogen is bonded to a more electronegative atom such as Oxygen, Chlorine and Fluorine.
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
The periodic table of elements arranges all of the known chemical elements in an informative array. Elements are arranged from left to right and top to bottom in order of increasing atomic number. Order generally coincides with increasing atomic mass. The rows are called periods.
