For NaBr(aq) hydrogen gas is formed at the cathode. Hydrogen in water is reduced to hydrogen gas. For NaBr(aq), bromine, Br2(l), will be produced instead of oxygen gas at the anode.
<span>For sodium sulfate hydrogen gas is formed at the cathode. Hydrogen in water is reduced to hydrogen gas. Oxygen gas will be produced at the anode. </span>
<span>Should someone suggest that sodium metal is formed at the cathode, rest assured that that can't happen. Sodium metal reacts with water to make Na+. </span>
<span>Lead(II) iodide is insoluble in water. Therefore, not much will happen since there is no electrolyte.</span>
Answer:
Explanation:the formula for osmotic pressure is given in the first line, all the terms are explained before solution, from the question,
Molar mass=Mm=386.6
Mass= m= 14.6
Volume= 263/1000 = 0.263dm^3
Temperature=T= 298K.
Gas constant .R= 0.082
Then substitute into the equation πV= nRT
Making π subject of formula
π= nRT/V
Then further substitution
Answer:
in the presence of water, a strong acid will dissociate quickly and release heat, so it is an exothermic reaction.
Explanation:
Answer:
Mixture of two elements
Explanation:
You can tell that it is an element since the atoms are only bonded to atoms of the same type. It is a mixture since the molecules are together in solution but aren't bonded together.
The additional volume of HCl which must be added to reach to the equivalence point is 8.33 mL
The moles of HCl which is required to reach the equivalence point can be calculated in the way as follows.
Moles of HCl can be calculated as
Moles of HCl = 0.004 moles of Ca (OH) 2 × 2 moles of HCl / 1 moles of Ca (OH) 2
= 0.008 moles of HCl
The volume of HCl which is required to reach the equivalence point can be calculated in the way given as follows.
Volume of HCl required= 0.008 moles of HCl × 1 L / 0.24 moles of HCl × 1 ml / 10 -³ L
= 33.33 ml
The additional volume of HCl calculated as
Additional volume = required volume – actual volume
= 33.33 mL – 25 mL
= 8 . 33 mL
Thus, we calculated that the additional volume of HCl which must be added to reach to the equivalence point is 8.33 mL.
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