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gogolik [260]
3 years ago
10

Each compound listed is soluble in water. For each compound, do you expect the resulting aqueous solution to conduct electrical

current? Why?
a) CsCl
b) CH3OH
c) Ca(NO2)2
d) C6H12O6
Chemistry
1 answer:
Strike441 [17]3 years ago
5 0

Answer:

See below in bold.

Explanation:

a) Yes  because it is an ionic compound. Forms ions in water solution.

b) This is an alcohol with covalent bonds it does not ionise in water - no.

c) Yes this forms ions in water solution.

d) This is a sugar and does not ionise so No.

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Carbon reacts with steam (h2o) at high temperatures to produce hydrogen and carbon monoxide. if 2 mol of carbon are exposed to 3
Ne4ueva [31]
The balanced equation for the above reaction is as follows;
C + H₂O ---> H₂ + CO
stoichiometry of C to H₂O is 1:1
1 mol of C reacts with 1 mol of H₂O
we need to find which is the limiting  reactant
2 mol of C and 3.1 mol of H₂O
therefore C is the limiting reactant and H₂O is in excess.
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4 0
3 years ago
When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant so
zmey [24]

Answer: 54.4 kJ/mol

Explanation:

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.0M\times 0.05=0.05mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.0\times 0.05L=0.05mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 mole

Now we have to calculate the mass of water:

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 50ml+50ml=100ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 100ml=100g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 100 g

T_{final} = final temperature of water = 27.5^0C

T_{initial} = initial temperature of metal = 21.0^0C

Now put all the given values in the above formula, we get:

q=100g\times 4.18J/g^oC\times (27.5-21.0)^0C

q=2719.6J=2.72kJ

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of HCl:

0.05 moles of HCl releases heat = 2.72 KJ

1 mole of HCl releases heat =\frac{2.72}{0.05}\times 1=54.4KJ

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ

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