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>
<span>HNO2 =====> H+ + NO2-
</span>I<span>nitial concentration</span> = 0.311
<span>C = -x,x,x </span>
<span>E = 0.311-x,x,x
</span>KNO2 ====>K+ + NO2-
<span>Initial concentration = 0.189 </span>
<span>C= -0.189,0.189,0.189 </span>
E = 0,0.189,0.189
Answer:
V ∝ n
Step-by-step explanation:
Suppose that pressure and temperature are constant.
If you try to force more molecules of air into a balloon, the balloon will expand.
This is an example of <em>Avogadro's Law</em>: the volume of a gas is directly proportional to the number of moles (particles).
V ∝ n
Answer:
Mass = 29.23 g
Explanation:
Given data:
Volume of solution = 814.2 mL 814.2/1000 = 0.8142 L)
Molarity of solution = 0.227 M
Mass of solute in gram = ?
Solution:
Molarity = number of moles / volume in L
By putting values,
0.227 M = number of moles / 0.8142 L
Number of moles = 0.227 M × 0.8142 L
Number of moles = 0.184 mol
Mass in gram:
Mass = number of moles × molar mass
Molar mass of calcium acetate = 158.17 g/mol
Mass = 0.184 mol × 158.17 g/mol
Mass = 29.23 g
Hey there!
Label A: Sublimation
Label B: Condensation
Label C: Melting
Remember sublimation is the transition of a substance directly from the solid to the gas state, without passing through the liquid state. Condensation is the conversion of a vapor or gas to a liquid. Melting is becoming liquefied by heat.
Hope this helps!
