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FrozenT [24]
3 years ago
6

If the standard solutions had unknowingly been made up to be 0.0024 m agno3 and 0.0040 m k2cro4, would this have affected your r

esults? how?
Chemistry
2 answers:
Mnenie [13.5K]3 years ago
6 0
2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with 
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04
OlgaM077 [116]3 years ago
6 0

If the standard solutions had unknowingly been made up to be 0.0024 M AgNO₃ and 0.0040 M K₂CrO₄ the chemical equation would be:  

2AgNO₃ + K₂CrO₄ → Ag₂CrO₄ (s) + 2KNO₃

By mixing 0.0024 M AgNO₃ and 0.004 M K₂CrO₄ one would have Ag₂CrO₄ precipitated out.  

Thus, one is left with 0.0024 M KNO₃ mixed with (0.004 -0.0024 / 2) M = 0.0028 M of K₂CrO₄.  

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Can someone please help me with this question also explain the answers I am so confused thank you.
Archy [21]

The theoretical yield of H₂S is 13.5 g.

The percent yield is 75.5 %.

<h3>What is the theoretical yield of H₂S from the reaction?</h3>

The equation of the reaction is given below:

  • FeS + 2 HCl → FeCl₂+ H₂S

Moles of FeS reacting = mass/molar mass

Molar mass of FeS = 88 g/mol

Moles of FeS reacting = 35/88 = 0.398 moles

Moles of H₂S produced = 0.398 moles

Molar mass of H₂S = 34 g/mol

Mass of H₂S produced = 0.398 * 34 = 13.5 g

Theoretical yield of H₂S is 13.5 g.

  • Percent yield = actual yield/theoretical yield * 100%

Actual yield of H₂S = 10.2 g

Percent yield = 10.2/13.5 * 100%

Percent yield = 75.5 %

In conclusion, the actual yield is less than the theoretical yield.

Learn more about percent yield at: brainly.com/question/8638404

#SPJ1

4 0
2 years ago
Differentiate between Mendeleev's Period law and Modern Periodic law<br>​
NeX [460]
<h2>see in the attachment hope it helps you</h2>

7 0
3 years ago
Calculate the mass defect for the formation of phosphorus-31. The mass of a phosphorus-31 nucleus is 30.973765 amu. The masses o
Nata [24]

<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399

<u>Explanation:</u>

Mass defect is defined as the difference in the mass of an isotope and its mass number.

The equation used to calculate mass defect follows:

\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M

where,

n_p = number of protons

m_p = mass of one proton

n_n = number of neutrons

m_n = mass of one neutron

M = mass number of element

We are given:

An isotope of phosphorus which is _{15}^{31}\textrm{P}

Number of protons = atomic number = 15

Number of neutrons = Mass number - atomic number = 31 - 15 = 16

Mass of proton = 1.00728 amu

Mass of neutron = 1.00866 amu

Mass number of phosphorus = 30.973765 amu

Putting values in above equation, we get:

\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399

Hence, the mass defect for the formation of phosphorus-31 is 0.27399

8 0
3 years ago
To convert measurements in the metric system you can move the decimal point left or right why does this work?
Serhud [2]

Explanation:

Because when you move the decimal point once, it it equal to 10 and to convert in the metric system, you must multiply or divide by 10. For example, when converting 4 centimeters to millimeters, you can multiply by 10 to get 40 millimeters or you can move the decimal point to the right and add another 0 to get 40.

3 0
3 years ago
Determine the oxidation number of the element "J" in H3JO2-
igor_vitrenko [27]

Answer:

Oxidation number:

3*1+ oxidation number of J+2*-2= -1

Oxidation number of J = 0

5 0
3 years ago
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