This question comes with four answer choices:
<span>A. H2O + H2O ⇄ 2H2 + O2
B. H2O + H2O⇄ H2O2 + H2
C. H2O + H2O ⇄ 4H+ + 2O2-
D. H2O + H2O ⇄ H3O+ + OH-
Answer: option </span><span>D. H2O + H2O ⇄ H3O+ + OH-
(the +sign next to H3O is a superscript, as well as the - sing next to OH)
Explanation:
The self-ionization of water, or autodissociation, produces the two ions H3O(+) and OH(-). The presence of ions is what explain the electrical conductivity of pure water.
</span><span>In this, one molecule of H2O loses a proton (H+) (deprotonates) to become a hydroxide ion, OH−. Then, he <span>hydrogen ion, H+</span>, immediately protonates another water molecule to form hydronium, H3O+.
</span>
Answer:
T₂ = 478.93 K
Explanation:
Given that,
Initial pressure, P₁ = 178 kPa
Initial temperature, T₁ = 37°C = 310 K
Final pressure, P₂ = 275 kPa
We need to find the new temp of the air in the tire, assuming the volume is constant. We know that,

So, the new temperature is equal to 478.93 K.
Answer:
A. DH° = –36 kJ
Explanation:
It is possible to obtain DH° of a reaction by the sum of DH° of half reactions. The DH° of the reaction:
B₂H₆(g) → 2B(s) + 3H₂(g)
Could be obtained from:
<em>(1) </em>2B(s) + 1.5O₂(g) → B₂O₃(s) DH° = –1273kJ
<em>(2) </em>B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g) DH° = –2035kJ
<em>(3) </em>H₂(g) + 0.5O₂(g) → H₂O(g) DH° = –242kJ
The sum of (2) - (1) gives:
B₂H₆(g) + 1.5O₂(g) → 2B(s) + 3H₂O(g) DH° = -2035kJ - (-1273kJ) = -762kJ
Now, this reaction - 3×(3):
B₂H₆(g) → 2B(s) + 3H₂(g) DH° = -762kJ - (3×-242kJ) = -36kJ
Thus, right answer is:
<em>A. DH° = –36 kJ</em>
Answer:
nếu có 40g dung dịch NaOH 20% phairn dùng hết bao nhiêu gam dung dịch HCl 25% để trung hoà
Answer:
21.344%
Explanation:
For the given chemical reaction, 8 moles of the reactant should produce 4 moles of
. However, 195 g of
was produced instead. The molar mass of
is 61.9789 g/mol.
Thus, the moles of
produced = 195/61.9789 = 3.1462 moles
The percent error = [(Actual -Experiment)/Actual]*100%
The percent error = [(4.00 - 3.1462)/4.00]*100% = (0.85376/4.00)*100% = 21.344%