Answer:
5.59 %
Explanation:
From the question given above, the following data were obtained:
Observed value of density = 2.85 g/cm³
True value of density = 2.699 g/cm³
Percentage error =.?
The percentage error of the student can be obtained as follow:
Percentage error = |Observed value – True value|/True value × 100
Percentage error = |2.85 – 2.699|/2.699
Percentage error = 0.151/2.699 × 100
Percentage error = 5.59 %
Therefore, the percentage error of the student is 5.59 %.
Answer:
48.37514 kj
Explanation:
Given data:
Mass of water = 163 g
Initial temperature = 29°C
Final temperature = 100°C
Heat added = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water is 4.18 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
ΔT = 100°C - 29°C
ΔT = 71°C
Q = 163 g × 4.18 j/g.°C × 71°C
Q = 48375.14 j
Joule to Kj conversion:
48375.14 /1000 = 48.37514 kj
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+.
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