Hi :)
20 mol NH3 x 6 H2O/4 NH3 = 30 mol H2O
Hope this helped :)
First, we need to compute the mass of oxygen found in 100 grams of saltpeter:
mass of oxygen = 100 - (mass of potassium + mass of nitrogen)
= 100 - (38.67 + 13.86)
= 100 - 52.53
mass of oxygen in 100 grams saltpeter = 47.47 grams
Now, we can use cross multiplication to find the mass of oxygen in 328 grams saltpeter as follows:
mass of oxygen = (328 x 47.47) / 100 = 155.7016 grams
Answer:
In one mole of glucose, there are
6.022×1023
individual glucose molecules
Explanation:
Hey there!
C₅H₅ + Fe → Fe(C₅H₅)₂
Put a coefficient of 2 in front of C₅H₅ on the left side because there is a subscript of 2 after C₅H₅ in parenthesis on the right.
2C₅H₅ + Fe → Fe(C₅H₅)₂
Fe (iron) is already balanced since there is one on each side, so we don't need to change anything for that.
This is a synthesis reaction because two reactants, C₅H₅ and Fe, are yielding a single product, Fe(C₅H₅)₂.
Hope this helps!
•3.9g of ammonia
•molar mass of ammonia = 17.03g/mol
1st you have to covert grams to moles by dividing the mass of ammonia with the molar mass:
(3.9 g)/ (17.03g/mol) = 0.22900763mols
Then convert the moles to molecules by multiplying it with Avogadro’s number:
Avogadro’s number: 6.022 x 10^23
0.22900763mols x (6.022 x 10^23 molecs/mol)
= 1.38 x 10^23 molecules
