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probablyacommunist
14.01.2021 •
Chemistry
is the base unit for length in the metric system
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Ответ:
metric system
Explanation:
In the metric system, the basic unit of length is the meter. A meter is slightly larger than a yardstick, or just over three feet. The basic metric unit of mass is the gram. Finally, the basic metric unit of volume is the liter.
Ответ:
After the reaction we have 5.46 grams of Ag2CO3, 3.37 grams of NaNO3 and remain 5.40 grams of Na2CO3
Explanation:
Step 1: Data given
A solution containing 7.50 g of sodium carbonate is mixed with one containing 6.75 g of silver nitrate.
Step 2: The balanced equation
Na2CO3 + 2 AgNO3 → Ag2CO3 + 2 NaNO3
Step 3: Calculate moles
Moles = mass / molar mass
moles Na2CO3 = 7.50 g/105.99 g/mol=0.0708 moles
moles AgNO3 = 6.75 g/ 169.87 g/mol= 0.0397 moles (limiting reactant)
Step 4: Calculate limiting reactant
For 1 mol Na2CO3 we need 2 moles AgNO3 to produce 1 mol Ag2CO3 and 2 moles NaNO3
AgNO3 is the limiting reactatn. It will completely be consumed. (0.0397 moles). Na2CO3 is in excess. There will react 0.0397/2 =0.01985 moles
There will remain 0.0708 - 0.01985 = 0.05095 moles Na2CO3
This is 0.05095 moles * 105.99 g/mol = 5.40 grams
Step 5: Calculate moles products
For 1 mol Na2CO3 we need 2 moles AgNO3 to produce 1 mol Ag2CO3 and 2 moles NaNO3
For 0.0397 moles AgNO3 we'll have 0.0397 moles NaNO3 and 0.01985 moles Ag2CO3
Step 6: Calculate mass
Mass Ag2CO3 = moles Ag2CO3 * molar mass
Mass Ag2CO3 = 0.01985 moles * 275.75 g/mol
Mass Ag2CO3 = 5.46 grams
Mass NaNO3 = 0.0397 moles * 84.99 g/mol
Mass NaNO3 = 3.37 grams
After the reaction we have 5.46 grams of Ag2CO3, 3.37 grams of NaNO3 and remain 5.40 grams of Na2CO3
7.50 + 6.75 grams = 14.25
Mass products = 5.46 + 3.37 grams = 8.83
mass of reactants in excess = 5.40 grams
remaining products = 8.83 + 5.40 = 14.23 grams