We are given with two reactants :barium hydroxide and sodium sulfate. The products of the reaction via double replacement is barium sulfate and sodium hydroxide. according to the solubility rules the product barium sulfate is insoluble. There is a one to one correspondence to every compound. In this case, we just have to find the limiting reactant and base the calculations there. The mass of the precipitate formed is 0.186 grams BaSO4
Answer:
5
Explanation:
chemical formulas show what atoms are in a molecule. In this case there is 1 hydrogen (H), 1 chlorine (Cl), and 3 oxygens (O). The 3 behind the oxygen is a subscript and tells us that there are 3 oxygen atoms.
Another example is H2O which as 3 atoms. 2 hydrogens (H) and 1 oxygen (O). This formula has a subscript 2 behind the hydrogen showing that there are 2 hydrogens.
Is that high school stuff or college?
Answer is: <span>the mass of the glucose is 81,07 grams.
</span>c(C₆H₁₂O₆) = 0,3 M = 0,3 mol/L.
V(C₆H₁₂O₆) = 1,500 L.
n(C₆H₁₂O₆) = c(C₆H₁₂O₆) · V(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 0,3 mol/L · 1,5 L.
n(C₆H₁₂O₆) = 0,45 mol.
m(C₆H₁₂O₆) = n(C₆H₁₂O₆) · M(C₆H₁₂O₆).
m(C₆H₁₂O₆) = 0,45 mol · 180,156 g/mol.
m(C₆H₁₂O₆) = 81,07 g.
Answer:
pH = 7.0
Explanation:
When HCl reacts with NaOH, H₂O and NaCl are produced, thus:
HCl + NaOH → H₂O + NaCl
At equivalence point, all HCl reacts with NaOH. The only you will have is water.
Equilbrium of water is:
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
K = 1x10⁻¹⁴ = [H⁺] [OH⁻]
As H⁺ = OH⁻ <em>because both are produced from the same water-</em>
1x10⁻¹⁴ = [H⁺]²
1x10⁻⁷M = [H⁺]
As pH = -log= [H⁺]
<h3>pH = 7.0</h3>
<em>-The pH at equivalence point in the titration of a strong acid with a strong base is always 7.0-</em>