The answer is 10 grams.
The atomic weights for each elements are :
<span>Na - 22.99 g/mol </span>
<span>O - 16.00 g/mol </span>
<span>H - 1.01 g/mol
The sum = 40 g/mol for NAOH
</span><span>0.250 moles * 40.00 g / 1 mole = 10 g NaOH</span>
Answer:
D is the answer because I think it is right plus I know they don't use two off them
They are able to be divided by a chemical reaction
To complete the balancing of the following combustion reaction, we must do elemental balances. For C, there is 5 on the left side hence there should be 5 CO2. For H, H20 should be 9/2. Next we balance O. on theright side we find a total of 29/2. So the o2 on the left side should be 29/4. To have whole number stoich.coeff. we multiply the numbers by 4, hence answer is
4C5H9O + 9O2= 20 O2 + 18 H20.
Answer:
a) pH will be 12.398
b) pH will be 4.82.
Explanation:
a) The moles of NaOH added = molarity X volume (L) = 2 X 0.01 = 0.02 moles
The total volume after addition of pure water = 0.780+0.01 = 0.79 L
The new concentration of /NaOH will be:
the [OH⁻] = 0.025
pOH = -log [OH⁻] = 1.602
pH = 14 -pOH = 12.398
b) The buffer has butanoic acid and butanoate ion.
i) Before addition of NaOH the pH will be calculated using Henderson Hassalbalch's equation:
![pH=pKa+log\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
pKa=
ii) on addition of base the pH will increase.
