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aniked [119]
3 years ago
13

Predict the products of the following reactions (assume complete reaction). In your submission include the overall balanced equa

tion, and balanced net ionic equation if applicable:
a. Adding zinc metal to a solution of copper(II) sulfate
b. Dripping silver perchlorate into a solution of sodium carbonate
c. A lump of aluminum metal is dropped into concentrated hydroiodic acid
d. Sucrose is fermented in water by yeast to make ethanol and carbon dioxide (note: water is a reactant)
e. Solid calcium oxide is added to water (careful)
f. A lump of potassium metal is dropped into water (careful)
Chemistry
1 answer:
olchik [2.2K]3 years ago
5 0

Answer:

Zn(s) + CuSO4(aq) -----> ZnSO4(aq) + Cu(s)

Na2CO3 (aq)+ 2AgClO4(aq) → 2NaClO4(aq) + Ag2CO3(s)

2Al(s) + 6HI(aq) ------> 2AlI3(aq) + 3H2(g)

C12H22O11(s)+H2O(l)→4C2H5OH(aq)+4CO2(g)

2K(s) + 2H2O(l) ------->2KOH(aq) + H2(g)

Explanation:

In the first reaction, zinc displaces copper from an aqueous copper II salt since zinc is higher than copper in the electrochemical series.

Reaction two leads to the precipitation of silver carbonate. Remember that carbonates are mostly insoluble in water.

Acids are known to displace hydrogen gas from dilute acids. Hence in the third reaction, aluminum displaced hydrogen from aqueous hydroiodic acid.

In the fourth reaction, the action of yeast leads to the fermentation of sucrose yielding ethanol and carbon dioxide in the process.

Metals are known to form aqueous alkaline solutions liberating hydrogen gas when they react with water. Hence potassium metal reacts with water to liberate hydrogen gas and form alkaline potassium hydroxide.

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How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o
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Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

Density (\rho)=\frac{Mass(m)}{Volume(V)}

Density of water = 1.0 g/mL

So, mass of water = 353 g

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Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,  

334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)

334x+x\times 11.976=29553.16

345.976x = 29553.16

x = 85.4197 kg

Thus,  

<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>

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