Answer: D) It conducts electricity when it is dissolved in water
The balanced equation for the acid base reaction is as follows
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
the number of NaOH moles reacted - 0.200 mol/L x 0.0250 L = 0.005 mol
according to molar ratio
number of NaOH moles reacted = number of HCl moles reacted
therefore number of HCl moles - 0.005 mol
volume of 30.0 mL contains 0.005 mol
therefore 1000 mL contains - 0.005 mol / 0.030 L = 0.167 M
concentration of HCl is 0.167 M
The solution to the problem is as follows:
<span>a)
C + O2 = CO2
Molar mass CO2 = 44 g/mol
3.67 g CO2 * 1 mol / 44 g =
=0.0834 mol CO2 = 0.0834 mol C
I hope my answer has come to your help. God bless and have a nice day ahead!
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Answer:
B 1.23 g/cc
Explanation:
For something to float on seawater, the density must be less than 1.03 g/mL. If the object sinks, the density is greater than 1.03 g/mL.
Let’s examine the answer choices. Keep in mind, the ice berg is mostly below the water level.
A. 0.88 g/cc
This is less than 1.03 g/cc, which would result in floating.
B. 1.23 g/cc
This is the best answer choice. The iceberg is mostly beneath the water, but some of it is exposed. The density is greater than 1.03 g/mL, but not so much greater that it would immediately sink.
C. 0.23 g/cc
This is less than 1.03 g/cc, which would produce floating.
D. 4.14 g/cc
This is much greater than 1.03 g/cc and the result would be sinking.
Answer:
1. The correct option is;
c. maintains charge balance in the cell
2. The correct option is;
c. +3.272 V
Explanation:
The aqueous solution in a galvanic cell is the electrolyte which is a ionic solution containing that permits the transfer of ions between the separated compartment of the galvanic cell such that the overall system is electrically neutral
Therefore, the aqueous solution maintains the charge balance in the cell
2. Here we have;
B₂ + 2e⁻ → 2B⁻ Ecell = 0.662 V
A⁺ + 1e⁻ → A Ecell = -1.305 V
Hence for the overall reaction, we have;
2A + B₂ → 2AB gives;
(0.662) - 2×(-1.305) = +3.272 V.