Element is a pure substance that is <span>made up of only one type of atom.
An element cannot be broken down by any chemical means. However, nuclear reactions can lead to the generation of new elements.
Examples of elements include: copper, neon and iron.
Currently, there are 115 known elements in nature.</span>
Answer:
1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
2) The amount (in grams) of excess reactant H₂ = 4.39 g.
Explanation:
- Firstly, we should write the balanced equation of the reaction:
<em>N₂ + 3H₂ → 2NH₃.</em>
<em>1) To determine the limiting reactant of the reaction:</em>
- From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- This means that <em>N₂ reacts with H₂ with a ratio of (1:3).</em>
- We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation:<em> n = mass / molar mass.</em>
The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.
The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.
- From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).
The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).
∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>
- As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
- Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
- The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.
- ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.
Answer:
Ionic bonds
Explanation:
It rymes. haha i dont even know how to spell it! ;)
They help scientists understand complex ideas and objects that aren’t easy to handle.
Explanation:
Normal moles of 
= volume × normal concentration
= 4.7 × 0.139 = 0.6533 mol
Moles of in hyponatremia blood = volume × hyponatremia concentration
= 4.7 × 0.116 = 0.5452 mol
Moles of NaCl to be added = moles of extra needed
= 0.6533 mol - 0.5452 mol = 0.1081 mol
Mass of NaCl = moles × molar mass of NaCl
= 0.1081 mol × 58.443
= 6.317g
= 6.32 g (approx)
Thus, we can conclude that mass of sodium chloride would need to be added to the blood is 6.32 g.
