Answer:
1) 10.0 moles of NO
2) 25 moles of NaCl
3) 1200 moles of CO2
1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in the following composition reaction?
N2 + O2 → 2NO
For 1 mol N2 we need 1 mol O2 to produce 2 moles of NO
For 5.0 moles of N2 we need 5.0 moles of O2 to produce 10.0 moles of NO
2. The neutralization of an acid with a base is a double replacement reaction in which a salt and water are formed. If you start with 25 moles of HCl and neutralize it with NaOH how many moles of NaCl will be formed?
HCl + NaOH → NaCl + H2O
For 1 mol HCl we need 1 mol NaOH to produce 1 mol of NaCl and 1 mol H2O
For 25 moles of HCl we need 25 moles of NaOH to produce 25 moles of NaCl and 25 moles of H2O
3. A car burns gasoline (octane – C8H18) with oxygen. If you drive to Salt Lake and burn 150 moles of octane how many moles of carbon dioxide are you producing?
2C8H18 + 25O2 → 16CO2 + 18H2O
For 2 moles of octane we need 25 moles of O2 to produce 16 moles of CO2 and 18 moles of H2O
For 150 moles of octane we need 25*75 = 1875 moles of O2
To produce 16*75 = 1200 moles of CO2 and 18*75= 1350 moles
Explanation:
Answer is: new substance.
For example, synthesis chemical reaction: Ba + F₂ → BaF₂.
Synthesis reaction is a type of reaction in which multiple reactants combine to form a single product.
New substance, barium fluoride is formed, with different chemical and ohysical properties than reactants (barium and fluorine).
In barium fluoride, barium has oxidation number +2 and fluorine has oxidation number -1, so compound has neutral charge.
Answer:
Average atomic mass of the vanadium = 50.9415 amu
Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975
Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu
Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025
Atomic mass of Isotope (II) of vanadium ,m' = ?
Average atomic mass of vanadium =
m × abundance of isotope(I) + m' × abundance of isotope (II)
50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025
m'= 49.944 amu
Explanation:
Your answer is c. Hoped this has helped the
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