How many gramsof sodium carbonate are neededto make1.30l of0.890m
Do you mean 0.890 m or 0.890 M?
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To determine the amount of sodium carbonate needed to make a solution of a certain molarity, you will need to use the formula:
Molarity = moles / volume
Given:
Molarity = 0.890 M
Volume = 1.30 L
First, we need to calculate the number of moles of sodium carbonate using the equation:
moles = Molarity * volume
Plugging in the values:
moles = 0.890 M * 1.30 L
Now, to calculate the mass of sodium carbonate, we need to use the equation:
mass = moles * molar mass
The molar mass of sodium carbonate (Na2CO3) is calculated by adding up the atomic masses of each element in the compound:
Molar mass of Na2CO3 = (2 * atomic mass of Na) + atomic mass of C + (3 * atomic mass of O)
Consulting the periodic table, we find:
Atomic mass of Na = 22.99 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of O = 16.00 g/mol
Now we can calculate the molar mass:
Molar mass of Na2CO3 = (2 * 22.99 g/mol) + 12.01 g/mol + (3 * 16.00 g/mol)
Finally, we can calculate the mass of sodium carbonate:
mass = moles * molar mass
Now you can substitute the values you have obtained into the equation to find the answer.