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2 years ago

Which of the following is considered a STRONG electrolyte? C12H22O11 HC2H3O2 PbCl2 CH3OH NH4NO3? please explain

1 answer:

6 0

I believe the correct answer from the choices listed above is the fifth option. Of the following , the strong electrolyte would be NH4NO3. NH4NO3<span> is a salt and completely dissociates in aqueous solution. Hope this answers the question. Have a nice day.</span>

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Which is the best interpretation of the two flat portions of the graph

Fed [463]

Answer:

Explanation:

5 0

2 years ago

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For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the

Aneli [31]

The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.

8 0

3 years ago

A _______ reaction is when a compound containing carbon and hydrogen (and sometimes oxygen) combines with oxygen gas to produce

ch4aika [34]

This is categorized as a combustion reaction.

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2 years ago

Are atoms of gold the same as antoms of oxygen? yes or no?

schepotkina [342]

Answer:

No they are not the same the are both di

8 0

2 years ago

A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 192.mg of oxalic acid H2

kap26 [50]

Answer:

The molarity of the NaOH solution is 0.076 M

Explanation:

Step 1: Data given

Mass of oxalic acid = 192 mg = 0.192 grams

volume = 250 mL = 0.250 L

Molar mass oxalic acid = 90.03 g/mol

Step 2: The balanced equation

H2C2O4 + 2NaOH → Na2C2O4 + H2O

Step 3: Calculate moles of oxalic acid

Moles oxalic acid = 0.192 grams / 90.03 g/mol

Moles oxalic acid = 0.00213 moles

Step 4: Calculate molarity of oxalic acid

Molarity = Moles / volume

Molarity = 0.00213 moles / 0.250 L

Molarity = 0.00852 M

Step 5: Calculate Molarity of NaOH

2 Ca*Va = Cb*Vb

with Ca = Molarity of oxalic acid = 0.00852 M

with Va = volume of oxalic acid = 0.250 L

with Cb = Molarity of NaOH = TO BE DETERMINED

with Vb = volume of 0.0558 L

Cb = (2*0.00852 * 0.25) / 0.0558

Cb = 0.076 M

The molarity of the NaOH solution is 0.076 M

4 0

2 years ago