Answer:
7.32g of HNO3 are required.
Explanation:
1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.
From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.
2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:
• starting with the 4.30 grams of Ca(OH)2.
,
• using the molar mass of Ca(OH)2 (74g/mol).
,
• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .
,
• using the molar mass of HNO3 (63.02g/mol).
So, 7.32g of HNO3 are required.
The combustion reaction of propane would be expressed as:
C3H8 + 5O2 = 3CO2 + 4H2O
To determine the mass of water that is produced from the given amount of propane, we use the mass of propane and the relation of the substances from the balanced reaction. We do as follows:
moles propane = 22 g C3H8 ( 1 mol / 44.1 g ) = 0.50 mol C3H8
moles H2O = 0.50 mol C3H8 ( 4 mol H2O / 1 mol C3H8) = 2 mol H2O
mass H2O = 2 mol H2O ( 18.02 g / 1 mol ) = 36.04 g H2O
Therefore, the mass of water that is produced from 22 grams of propane would be 36.04 g.
Answer:
24 sultur with 48hydrogen+16 hydrogen,nitrogen and 48 oxygen+16 nitrogen,16 oxygen+62hydrogen and 32 oxygen
Explanation:
The length of the seasons would increase
