Answer:
0.129 L = 129.0 mL.
Explanation:
- NaOH neutralizes acetic acid (CH₃COOH) according to the balanced reaction:
<em>NaOH + CH₃COOH → CH₃COONa + H₂O. </em>
- According to the balanced equation: 1.0 mole of NaOH will neutralize 1.0 mole of CH₃COOH.
<em>no. of moles of CH₃COOH = mass/molar mass </em>= (2.0 g)/(60 g/mol) = <em>0.033 mole. </em>
<em>
</em>
no. of moles = (0.258 mol/L)(V)
- At neutralization: no. of moles of NaOH = no. of moles of CH₃COOH
∴ (0.258 mol/L)(V) = 0.033 mole
<em>∴ The volume of NaOH</em> = (0.033 mole)/(0.258 mol/L) = <em>0.129 L = 129.0 mL.</em>
Answer: Option A) 2 atoms
Explanation:
4Al + 302 ---> 2Al2O3
4 moles of Al reacts with 3 molecules of O2
So 1.10 moles of Al should reacts with Z molecules of O2 (Assume Z is unknown)
i.e 4 moles of Al = 3 molecules of O2
4 moles of Al = 6 atoms of O2 (since 1 molecule of oxygen has two atoms of oxygen)
1.10 atoms of Al = Z atoms of O2
To get the value of Z, cross multiply
Z x 4 moles = 6 atoms x 1.10 moles
4Z = 6.6
Divide both sides by 4
4Z/4 = 6.6/4
Z = 1.65 atoms (approximately 2 atoms)
Thus, 2 atoms of the excess reactant oxygen remains.
Answer:
pH = 2.87
Explanation:
Let us represent uric acid as HA
The dissociation will be:
Let "x" concentration of HA dissociates so it will give x concentration of hydrogen ion and the conjugate base.
![Ka=\frac{[H^{+}][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
pKa = 3.89
Therefore Ka = 0.000129
Putting values
on may ignore x in denominator
x = 0.00134 M
[H⁺] = 0.00134
pH = -log[H⁺] = 2.87
Explanation:
Al2O3
16=molar mass of one oxygen atom
Number of moles: mass/molar mass
Mol of O3: 19÷(16*3) = 0.40 g/mol
mole ratio
0.40: 3 (oxygen)
0.27: 2 (aluminum)
Mass = molar mass*mole
0.27*(27*3)= 21.87g
