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3 years ago

If an experiment produces 5 g but should have made 500 g, what is the percent yield?

Chemistry

1 answer:

hoa [83]3 years ago

7 0

Answer:

Percentage of yield = 1%

Explanation:

Given:

Amount of yield = 5g

Total amount of product = 500 gram

Find:

Percentage of yield = ?

Computation:

⇒ Percentage of yield = [Amount of yield / Total amount of product]100

⇒ Percentage of yield = [5g / 500g]100

⇒ Percentage of yield = [0.01]100

⇒ Percentage of yield = 1%

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When 0.200 grams of Al reacts with 15.00 mL of a 0.500 M copper(II) chloride solution, how many moles of solid Cu would be produ

Naddik [55]

When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)

where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute:
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025= 0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
= 0.0075 * 63.546 =0.477 g

3 0

3 years ago

You are a researcher for a golf club manufacturer. You are given two identical looking cubes of a metal alloy. You are informed

Rashid [163]

Answer:

C.Melt both cubes and look for a broader range of melting temperatures. The one that melts over a broader range of temperatures is the amorphous solid.

Explanation:

Amorphous solids is one that do not have a fixed melting points but melt over a wide range of temperature due to the irregular shape hence its name. Contrariwise crystalline solids, have a fixed and sharp melting point.

This comes in handy to solve the riddle. We can characterise the pair with the melting point property.

7 0

3 years ago

Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions.

sammy [17]

A electrochemical reaction is said to be spontaneous, if $E^{0} cell is positive.$

Answer 1:
Consider reaction: Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s)

The cell representation of above reaction is given by;
 $S^{2-}/S // Ni^{2+}/Ni$

Hence, $E^{0}cell = E^{0} Ni^{2+/Ni} - E^{0} S/S^{2-}$
we know that, ${E^{0} Ni^{2+}/Ni = -0.25 v$
and ${E^{0} S/ S^{2-} = -0.47 v$

Therefore, $E^{0} cell$ = - 0.25 - (-0.47) = 0.22 v

Since, $E^{0} cell$ is positive, hence cell reaction is spontaneous
.....................................................................................................................

Answer 2:
Consider reaction: Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)

The cell representation of above reaction is given by;
 $H_{2} / H^{+} // Pb^{2+} /Pb$

Hence, $E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+}$
we know that, ${E^{0} Pb^{2+}/Pb = -0.126 v$
and ${E^{0} H_{2}/ H^{+} = -0 v$

Therefore, $E^{0} cell$ = - 0.126 - 0 = -0.126 v

Since, $E^{0} cell$ is negative, hence cell reaction is non-spontaneous.

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Answer 3:
Consider reaction: 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)

The cell representation of above reaction is given by;
 $Cr/Cr^{2+} // Ag^{+}/Ag$

Hence, $E^{0}cell = E^{0} Ag^{+}/Ag - E^{0} Cr/Cr^{2+}$
we know that, ${E^{0} Ag^{+}/Ag = -0.22 v$
and ${E^{0} Cr/ Cr^{2+} = -0.913 v$

Therefore, $E^{0} cell$ = - 0.22 - (-0.913) = 0.693 v

Since, $E^{0} cell$ is positive, hence cell reaction is spontaneous

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Neporo4naja [7]

Answer:

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Explanation:

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Answer:

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