Hello!
We have the following data:
MM (Molar mass of Na2CO3)
Na = 2*23 = 46 u
C = 1*12 = 12 u
O = 3*16 = 48 u
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MM (Molar mass of Na2CO3) = 46+12+48 = 106 g/mol
n (mol number) = 1 mol
m1 (mass of the solute) = ?




So if we have the dissolution in a liter (m2 - mass of solvent), that is 1000g , then the total mass of the solution (m) will be:



<span>Now, let's find the percentage in mass (% m / m), let's see:
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Answer:
B) 9.6I Hope this helps, greetings ... DexteR!
Answer:
KE = 1/2 mv^2=0.5x 10 X 20^2= 800
Explanation
Answer:
See explanation
Explanation:
If we look at the models, we will see that the three fluorine atoms in CF3COOH are attached to the carbon that is next to the -COOH group.
As a result of the electron withdrawing effect of the three fluorine atoms, CF3COOH is much more acidic (104 times more acidic) than CH3COOH. This is reflected in the value of the Ka for each acid.
This electron withdrawing effect of the three fluorine atoms also stabilizes CF3COO- much more than CH3COO-.
Answer:
C2H6 + 7/2O2 ....... 2CO2 + 3H2O