Mars........................
Answer:
Violet is my favorite colour i think its so pretty
Explanation:
Your so nice
Answer:
0.56 g
Explanation:
<em>A chemist determines by measurements that 0.020 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrogen gas that participates.</em>
Step 1: Given data
Moles of nitrogen gas (n): 0.020 mol
Step 2: Calculate the molar mass (M) of nitrogen gas
Molecular nitrogen is a gas formed by diatomic molecules, whose chemical formula is N₂. Its molar mass is:
M(N₂) = 2 × M(N) = 2 × 14.01 g/mol = 28.02 g/mol
Step 3: Calculate the mass (m) corresponding to 0 0.020 moles of nitrogen gas
We will use the following expression.
m = n × M
m = 0.020 mol × 28.02 g/mol
m = 0.56 g
44g of CO2 can produce by the reaction of carbon with oxygen
Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
