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2 years ago

Which of the following

1 answer:

7 0

A. Acids and bases don't react with each other. --> False. Acids and bases react in reactions called neutralization reactions, whose products are salt and water.

B. Acids mixed with bases neutralize each other. --> True. See above.

C. Acids mixed with bases make stronger bases. --> False. Acids and bases react to neutralize each other, so they would weaken each other, not result in a stronger base or a stronger acid.

D. Acids mixed with bases make stronger acids. --> False. See above.

<h3>Answer:</h3>

B. Acids mixed with bases neutralize each other

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Complete and balance the molecular equation for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(N

Rina8888 [55]

Answer:

1. The balanced molecular equation is given below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

2. The net ionic equation is given below:

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

Explanation:

1. The balanced molecular equation

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

The above equation can be balance as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

Now, the equation is balanced.

2. The bal net ionic equation.

This can be obtained as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —>

In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:

Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)

Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)

Na2SO4(aq) + Ba(NO3)2(aq) —>

2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)

Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

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3 years ago

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3 years ago

2Al(s)+Fe2O3(s) → Al2O(s)+2Fe(s) with a delta H=-847 KJ.

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∆H for given reaction -847kJ

As it's negative reaction is exothermic

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1694KJ

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2 years ago

A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol

Scorpion4ik [409]

Answer:

see explanation

Explanation:

Step 1: Data given

Volume of 1M CusO4 = 50.0 mL = 0.05 L

Volume of 2M KOH = 50.0 mL = 0.05 L

Temperature before mixing= 17.8 °C

Temperature after mixing = 32.4 °C

The heat capacity of the calorimeter is 12.1 J/K

Step 2: The balanced equation

CuSO4(aq)+ 2KOH(aq) →Cu(OH)2(s) + K2SO4(aq)

Step 3: Calculate mass of the solution

Suppose the density of the solution is 1 g/mL

Total volume = 100 mL

Mass of the solution = density * volume

Mass of the solution = 1g/mL * 100 mL = 100 grams

Step 4:

Q = m*c*ΔT

with m = the mass of the solution = 100 grams

with c= the heat capacity of the solution = 4.184 J/g°C

with ΔT = 32.4 - 17. 8 = 14.6 °C

Q = 6108.64 J

Step 5: Calculate the energy of the calorimeter

Q = c*ΔT

Q = 12.1 J/K * 14.6

Q = 176.66 J

Step 6: Calculate total heat

Qtotal = 6108.64 + 176.66 = 6285.3 J = 6.29 kJ (negative because it's exothermic)

Step 7: Calculate moles

Moles CuSO4 = 0.05 L * 1M = 0.05 moles

Moles KOH = 0.05 L * 2M = 0.10 moles

ΔH = -6.29 kJ / 0.05 moles = -125.8 kJ/mol

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