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kherson [118]
3 years ago
7

A certain reaction has a 73.6% yield. if 53.8 grams of the product were predicted by stoichiometry to be made, what would the ac

tual yield be?
Chemistry
1 answer:
krok68 [10]3 years ago
5 0
Actual yield and theoretical yield are usually different.
the percentage yield of product = actual yield / theoretical yield x 100%
the theoretical yield is 53.8 g 
and percentage yield - 73.6 %
from the theoretical yield only 73.6 % of that amount is in the actual yield 
therefore actual yield is 73.6 % of 53.8 g
actual yield - 53.8 x 73.6% = 39.6 g
actual yield is 39.6 g
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weqwewe [10]

Answer:

Option D: it's ability to lose electrons

Explanation:

Alkali metals are usually discovered in nature. They have highly reactivity at STP conditions (standard temperature and pressure conditions) and easily lose their outermost electron to form positive ions known that have a charge of +1.

Thus, what can determine the extent of reactivity of an alkali metal, is it's ability to lose electrons

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2 years ago
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IrinaVladis [17]
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3 years ago
Given the following reaction: \ce{Cu + 2AgNO3 -> 2Ag + Cu(NO3)2}Cu+2AgNO3 ​ ​ 2Ag+Cu(NOX 3 ​ )X 2 ​ How many moles of \ce{Ag}
Alexandra [31]

Answer:

0.252 mol

Explanation:

<em>Given the following reaction: </em>

<em>Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂</em>

<em>How many moles of Ag will be produced from 16.0 g Cu, assuming AgNO3 ​ is available in excess.</em>

First, we write the balanced equation.

Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂

We can establish the following relations.

  • The molar mass of Cu is 63.55 g/mol.
  • The molar ratio of Cu to Ag is 1:1.

The moles of Ag produced from 16.0 g of Cu are:

16.0gCu.\frac{1molCu}{63.55gCu} .\frac{1molAg}{1molCu} =0.252 molAg

6 0
3 years ago
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An unknown element sample has 2 isotopes present. The first isotope has a mass of 6.017 amu and is
oksano4ka [1.4K]

Answer:

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Change each percent abundance into decimal form by dividing by 100. Multiply this value by the atomic mass of that isotope. Add together for each isotope to get the average atomic mass.

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8 0
2 years ago
How many grams of Pb are contained in a mixture of 0.135 kg each of PbCl(OH) and Pb2Cl2CO3?
Dahasolnce [82]

Hey there!:

Given the mass of PbCl(OH) :

0.135 Kg = 0.135 Kg*(1000g / 1Kg)  = 135 g

Molecular mass of PbCl(OH)  = 207+35.5+16+1  = 259.5 g / mol

Atomic mass of Pb = 207 g/mol

Hence mass of Pb in 135 g  PbCl(OH)  :

(207 g Pb /  259.5 g PbClOH) * 135g PbClOH  =

0.79768 * 135 =>  107.68 g of Pb

For Pb2Cl2CO3  :

Given the mass of Pb2Cl2CO3  :

0.135 Kg = 0.135 Kgx(1000g / 1Kg)  = 135 g

Molecular mass of Pb2Cl2CO3  = 2*207+2*35.5+12+3*16  = 545 g / mol

Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3  = 2*207 = 414 g

Hence mass of Pb in 135 g  Pb2Cl2CO3:

(414 g Pb /  545 g PbClOH) * 135g PbClOH  =

0.75963 * 135 => 102.55 g of Pb2Cl2CO3


Hope that helps!

8 0
3 years ago
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