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Snowcat [4.5K]
3 years ago
11

14. If 7.00 mol of PCI, and 5.00 mol of H2O react in the following chemical reaction, wind

Chemistry
1 answer:
Ymorist [56]3 years ago
6 0

Answer:

Option A. H₂O

Explanation:

PCl3 + 3H2O → H3PO3 + 3HCI

From the equation,

We can see clearly that for every 3moles of H2O, 1mole of PbCl3 is needed. This clearly indicates that H2O is the limiting reactant. This can further be proved by doing the following:

From the equation,

1mole of PbCl3 required 3moles of H20.

Therefore, 7moles of PbCl3 will require = 7x3 = 21moles of H2O. This amount(i.e 21moles) of H20 is far greater than what was given (i.e 5moles of H2O) from the question.

Now let us consider the reverse case as follows:

From the equation,

1mole of PbCl3 required 3moles of H20.

Therefore, Xmol of PbCl3 will require 5moles of H2O i.e

Xmol of PbCl3 = 5/3 = 1.67moles

This amount(i.e 1.67moles of PbCl3) obtained is smaller than what was given (i.e 7moles of PbCl3) from the question.

This shows that PbCl3 is excess and H20 is the limiting reactant

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If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0
2 years ago
You have 4 litres of a 3.0 mol/L solution of NaCl in a chemical store room.
ahrayia [7]

Answer:

12

Explanation:

nNaCl= 4x3=12

3 0
3 years ago
Why don't siblings look exactly alike
marusya05 [52]

Answer:

Your genes play a big role in making you who you are. ... But brothers and sisters don't look exactly alike because everyone (including parents) actually has two copies of most of their genes. And these copies can be different. Parents pass one of their two copies of each of their genes to their kids.

3 0
3 years ago
What is the density of an object with 525 grams and a volume of 15cm3?
Hunter-Best [27]

Answer: 35 g/cm

Explanation:

Density equals mass over volume. 525 divided by 15 is 35

7 0
3 years ago
Read 2 more answers
A scientist prepared an aqueous solution of a 0.45 M weak acid. The pH of the solution was 2.72. What is the percentage ionizati
aleksandr82 [10.1K]

Answer:

0.42%

Explanation:

<em>∵ pH = - log[H⁺].</em>

2.72 = - log[H⁺]

∴ [H⁺] = 1.905 x 10⁻³.

<em>∵ [H⁺] = √Ka.C</em>

∴ [H⁺]² = Ka.C

∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.

<em>∵ Ka = α²C.</em>

Where, α is the degree of dissociation.

<em>∴ α = √(Ka/C) </em>= √(8.065 x 10⁻⁶/0.45) = <em>4.234 x 10⁻³.</em>

<em>∴ percentage ionization of the acid = α x 100</em> = (4.233 x 10⁻³)(100) = <em>0.4233% ≅ 0.42%.</em>

4 0
3 years ago
Read 2 more answers
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