False, in converting between units, it is never necessary to use more than one conversion factor.
Answer is: 2) 117g.
2Na + Cl₂ → 2NaCl
Step 1: calculate amount of substance of sodium and chlorine.
n(Na) = m(Na)÷M(Na) = 46g ÷ 23 g/mol = 2 mol.
n(Cl₂) = m(Cl₂)÷M(Cl₂) = 71g ÷ 71 g/mol = 1 mol.
Step 2: calculate amount of substance and mass of sodium-chloride.
Because both sodium and chlorine react completely, we can use both n to compare with n of NaCl.
n(Na) : n(NaCl) = 2:2, 2 mol : n(NaCl) = 2:2
n(NaCl) = 2mol, m(NaCl) = 2mol ·5805 g/mol = 117 g.
Answer:
B)Continents look like they fit together
Answer:
Nitrogen is limiting reactant while hydrogen is in excess.
Explanation:
Given data:
Mass of N₂ = 25 g
Mass of H₂ = 25 g
Mass of ammonia formed = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of Nitrogen:
Number of moles = mass/ molar mass
Number of moles = 25 g/ 28 g/mol
Number of moles = 0.89 mol
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 25 g/ 2 g/mol
Number of moles = 12.5 mol
Now we will compare the moles of both reactant with ammonia.
H₂ ; NH₃
3 : 2
12.5 : 2/3×12.5 = 8.3
N₂ ; NH₃
1 : 2
0.89 : 2×0.89 = 1.78
The number of moles of ammonia produced by nitrogen are less thus nitrogen is limiting reactant while hydrogen is in excess.
Answer:
Option 4 ) 1-butyne
Explanation:
In organic chemistry, you should use IUPAC convention in order to name an organic compound. First of all, you should identify the lenght of the organic chain, for this case, you have 5 C atoms, but in fact, you have a triple bond (which would be a substitute: ethynil-) as a substitute, so the main skeleton would have 4 C atoms (a butane)
Then, you start by numbering carbon N° 1 as the one that has the substitute (triple bound)-starting from the right, it would be the second C):
CH₃-CH₂-CH₂-C≡CH
Which will finally leads us to 1-butyne
