0
4.828gNa X (1 mole/22.989gNa) = .21
22.989 = atomic weight of Na
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How many moles are present in 585 g of sodium chloride?
There are 10 moles present in 585 g of sodium chloride.
How many moles of NaCl are there in crystal of halite that weight is 5.88 g?
The answer is 0,1014 moles of sodium.
How many moles of sodium sulfate are present in 284 g?
2
In the formula for sodium phosphate Na3PO4 how many moles of sodium are represented?
the formula does not show how many moles are in a compound, but it tells how many atoms of an element that are in the compound...
How many sodium atoms are present in 0.2310 g of sodium?
When the RAM of sodium is assumed as 23.1, the given mass contains 0.01 moles of the substance.
How many moles are present in 585 g of sodium chloride?
There are 10 moles present in 585 g of sodium chloride.
How many grams are in 55.0 moles of sodium?
55.0 moles of sodium are equal to 1 264,44 g.
How many moles of Na2SO3 in 1 g?
1 g of sodium sulfite is equivalent to 0,0079 moles.
How many grams are in 0.40 moles of sodium chloride?
0,40 moles of sodium chloride contain 23,376 g.
How many moles are in 46.0 g of sodium?
Atomic Weight sodium = 23 so 46/23 = 2 moles of sodium in 46.0g
Which is heavier 5 moles of carbon or 5 moles of sodium?
5 moles of carbon is 60,055 g.5 moles of sodium is 114,949 g.
How many moles of NaCl are there in crystal of halite that weight is 5.88 g?
The answer is 0,1014 moles of sodium.
How many moles of sodium sulfate are present in 284 g?
2
In the formula for sodium phosphate Na3PO4 how many moles of sodium are represented?
the formula does not show how many moles are in a compound, but it tells how many atoms of an element that are in the compound...
How many moles are in 4.99 g of Na?
0.21705 moles. There are 22.9898 grams sodium in 1 mole.
How many grams of NaCl are present in 10.00 moles?
10 moles of sodium chloride have 584,397 g.
How many moles of sodium hydroxide will be produced from the complete reaction of 15.0 g sodium?
Sodium reacts with water. 0.652 NaOH moles will form.
