The grams of Na that are needed to complete to react with 40..0 g of O2 is calculated as below
find the moles of O2 used = mass/molar mass
= 40 g/32g/mol = 1.25 moles
write the reacting equation
4Na+ O2 = 2Na2O
by use of mole ratio between Na to O2 which is 4 :1
the moles of Na = 1.25 x 4 = 5 moles
mass of Na = mass x molar mass
= 5 moles x 23 g /mol= 115 moles
Answer:
The molarity of the solution is 1, 23 M.
Explanation:
We calculate the weight of 1 mol of NaCl from the atomic weights of each element of the periodic table. Then, we calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter)
Weight 1 mol NaCl= Weight Na + Weight Cl= 23 g + 35, 5 g= 58, 5 g
58, 5 g-----1 mol NaCl
89,94 g ---------x= (89,94 g x 1 mol NaCl)/58, 5 g= 1,54 mol NaCl
1249 ml solution------ 1,54 mol NaCl
1000ml solution------x= (1000ml solutionx 1,54 mol NaCl)/1249 ml solution
x=1,23 mol NaCl---> The solution is 1, 23 molar (1,23 M)
Answer: A hydrate is found to have the following percent composition: 48.8% MgSO4 and 51.2% water.
Explanation:
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