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kakasveta [241]
3 years ago
15

The percentage yield for the reaction

Chemistry
1 answer:
makkiz [27]3 years ago
5 0

Answer:

92.87 g.

Explanation:

∵ The percentage yield = (actual yield/theoretical yield)*100.

  • We need to calculate the theoretical yield:

From the balanced reaction:

<em>PCl₃ + Cl₂ → PCl₅,</em>

It is clear that 1 mol of PCl₃ reacts with 1 mol of Cl₂ to produce 1 mol of PCl₅.

  • We need to calculate the no. of moles of 73.7 g PCl₃:

n = mass/molar mass = (73.7 g)/(137.33 g/mol) = 0.536 mol.

<u><em>Using cross multiplication:</em></u>

1 mol of PCl₃ produce  → 1 mol of PCl₅, from stichiometry.

∴ 0.536 mol of PCl₃ produce  → 0.536 mol of PCl₅.

∴ The mass of PCl₅ (theoretical yield) = (no. of moles) * (molar mass) = (0.536 mol)*(208.24 g/mol) = 111.62 g.

<em>∵ The percentage yield = (actual yield/theoretical yield)*100.</em>

The percentage yield = 83.2%, theoretical yield = 111.62 g.

∴ The actual yield of PCl₅ = (The percentage yield)(theoretical yield)/100 = (83.2%)(111.62 g)/100 = 92.87 g.

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shtirl [24]

Answer : The heat your body transfer must be, 25.1 kJ

Explanation :

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat = ?

m = mass of water = 500.0 g

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

T_1 = initial temperature  = 25.0^oC

T_2 = final temperature  = 37.0^oC

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

Q=500.0g\times 4.18J/g^oC\times (37.0-25.0)K

Q=25080J=25.1kJ

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3 0
3 years ago
A strontium-90 atom that has a lost two electrons has __________ protons, __________ neutrons, and __________ electrons.
MAXImum [283]
<span><em>Answer:</em>
A strontium-90 atom that has a lost two electrons has <u>38</u> protons, <u>52</u> neutrons, and <u>36</u> electrons.

<em>Explanation:
</em>Atomic number<em> of </em>Strontium (Sr) is 38. 
<em>Atomic number = number of protons 
</em>Hence, Strontium has 38 protons.

If the element is in neutral state,
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The given mass number is 90. Hence, number of neutrons should be 90 - 38 = 52.</span>
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Carboxylic acid and esters can be easily distinguished on the basis of IR as carboxylic acid would contain a broad intense peak in 2500-3200cm_{-1} corresponding to OH stretching frequency whereas esters would not contain any such broad intense peak.

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The oxygen of the carbonyl group is protonated using the acidic proton which  leads to the generation of positive charge on the oxygen. The positive charge generated is delocalised over the whole acid molecule and hence the electrophilicity of carbonyl group is increased. Kindly refer attachment for the structures.

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