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xxTIMURxx [149]
3 years ago
5

2. Percent recovery and percent yield are occasionally used as if they are interchangeable terms. They do, however, have differe

nt meanings. Why is it more correct to use percent recovery here?
Chemistry
1 answer:
Mnenie [13.5K]3 years ago
4 0

Answer:

Answer is in the explanation.

Explanation:

The principal difference between percent recovery and percent yield is that the second involves a chemical reaction whereas the first is more about a purification process.

Thus, percent yield is defined as the ratio between amount of product and theoretical yield.

Percent recovery is the ratio between final amount of purified substance and initial amount of substance.

I hope it helps!

You might be interested in
An aqueous solution of potassium carbonate combine with a solution of calcium nitrate. What are the total and net ionic equation
romanna [79]
<span><span>K_2</span>C<span>O_3</span>(aq)+Ca(N<span>O_3</span><span>)_2</span>(aq)→ ?</span>
If we break these two reactants up into their respective ions, we get...<span><span>
K^+ </span>+ C<span>O^2_3 </span>+ C<span>a^<span>2+ </span></span>+ N<span>O_−3</span></span>
If we combine the anion of one reactant with the cation of the other and vice-versa, we get...<span>
CaC<span>O_3 </span>+ KN<span>O_3</span></span>

Now we need to ask ourselves if either of these is soluble in water. Based on solubility rules, we know that all nitrates are soluble, so the potassium nitrate is. Alternatively, we know that all carbonates are insoluble except those of sodium, potassium, and ammonium; therefore, this calcium carbonate is insoluble. This is good. It means we have a driving force for the reaction! That driving force is that a precipitate will form. In such a case, a precipitation reaction will occur, and the total equation will be...<span><span>
K_2</span>C<span>O_3</span>(aq) + Ca(N<span>O_3</span><span>)_2</span>(aq) → CaC<span>O_3</span>(s) + 2KN<span>O_3</span>(aq)</span>

To determine the net ionic equation, we need to remove all ions that appear on both sides of the equation in aqueous solution -- these ions are called spectator ions, and do not actually undergo any chemical reaction. To determine the net ionic equation, let's first rewrite the equation in terms of ions...
2K^+(aq) + CO_3^{2-}(aq) + Ca^{2+}(aq) + 2NO_3^{-}(aq) → Ca^{2+}(s) + CO_3^{2-}(s) + 2K^+(aq) + 2NO_3^-(aq)

The species that appear in aqueous solution on both sides of the equation (spectator ions) are... <span>
2K^+,NO_3^-</span>
If we remove these spectator ions from the total equation, we will get the net ionic equation...
CO_3^{2-}(aq) + Ca^{2+}(aq) <span>→</span> CaCO_3(s)
5 0
3 years ago
What volume of ammonia gas, measured at 547.9 mmHg and 27.6oC, is required to produce 8.98 g of ammonium sulfate according to th
Studentka2010 [4]

Answer:

4.65 L of NH₃ is required for the reaction

Explanation:

2NH₃(g)  +  H₂SO₄(aq)  → (NH₄)₂SO₄(s)

We determine the ammonium sulfate's moles that have been formed.

8.98 g . 1mol / 132.06 g = 0.068 moles

Now, we propose this rule of three:

1 mol of ammonium sulfate can be produced by 2 moles of ammonia

Therefore, 0.068 moles of salt were produced by (0.068 . 29) / 1 = 0.136 moles of NH₃. We apply the Ideal Gases Law, to determine the volume.

Firstly we do unit's conversions:

27.6°C +273 =  300.6 K

547.9 mmHg . 1 atm / 760 mmHg = 0.721 atm

V = ( n . R . T ) / P → (0.136 mol . 0.082 L.atm/mol.K . 300.6K) / 0.721 atm

V = 4.65 L

4 0
3 years ago
Read 2 more answers
Please help!!!!! I need the correct answer quickly!!!
xxMikexx [17]

Answer :

The oxidation state of oxygen (O) in OF_2  is, (+2)

The oxidation state of carbon (C) in CO  is, (+2)

The oxidation state of nitrogen (N) in K_3N  is, (-3)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of  Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of  oxygen (O)  in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

(a) The given compound is, OF_2

Let the oxidation state of 'O' be, 'x'

x+2(-1)=0\\\\x-2=0\\\\x=+2

The oxidation state of oxygen (O) in OF_2  is, (+2)

(b) The given compound is, CO

Let the oxidation state of 'C' be, 'x'

x+(-2)=0\\\\x-2=0\\\\x=+2

The oxidation state of carbon (C) in CO  is, (+2)

(c) The given compound is, K_3N

Let the oxidation state of 'N' be, 'x'

3(+1)+x=0\\\\3+x=0\\\\x=-3

The oxidation state of nitrogen (N) in K_3N  is, (-3)

7 0
3 years ago
Read 2 more answers
Using the periodic table entry below, match the phrases with their corresponding values.
zhannawk [14.2K]

Atomic mass = 39

number of neutrons = 20

Number of protons= 19

Electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

The number of electron in first shell = 2

number of electrons in n=3 shell = 8


5 0
3 years ago
Read 2 more answers
What volume of 1.75 M hydrochloric acid (HCl aq) must be diluted with water to prepare 0.500 L of 0.250 M hydrochloric acid?
Airida [17]

Answer:

V_1=0.0714L

Explanation:

Hello there!

In this case, since we need to dilute the 1.75-M HCl, and the total number of moles remain unchanged, we can write:

n_1=n_2

And in terms of volume and concentration:

C_1V_1=C_2V_2

Thus, we can solve for the volume of the concentrated HCl as shown below:

V_1=\frac{C_2V_2}{C_1}

Therefore, we plug in the data to get:

V_1=\frac{0.250M*0.500L}{1.75 M}\\\\V_1=0.0714L

Best regards!

7 0
3 years ago
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