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stich3 [128]
3 years ago
10

Give another mixture of liquids that is separated on an industrial scale by fractional

Chemistry
1 answer:
iren [92.7K]3 years ago
7 0

Answer:

dna or diabeties can be separated

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Which of the following pairs lists a substance that can neutralize H2SO4 and the salt that would be produced from the reaction?
Bogdan [553]

The second option only.

  • LiOH, Li₂SO₄.
<h3>Explanation</h3>

A base neutralizes an acid when the two reacts to produce water and a salt.

Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:

Metal hydroxides

  • Ca(OH)₂ and
  • LiOH.

Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.

\text{Ca}(\text{OH})_{\bf 2}+\text{H}_2\text{SO}_4 \to \textbf{Ca}\textbf{SO}_{\bf 4} +{\bf 2}\;\text{H}_2\text{O}.

The formula for calcium sulfate \text{CaSO}_4 in option A is spelled incorrectly. Why? The charge on each calcium \text{Ca}^{2+} is +2. The charge on each sulfate ion {\text{SO}_4}^{2-} is -2. Unlike \text{Li}^{+} ions, it takes only one \text{Ca}^{2+} ion to balance the charge on each {\text{SO}_4}^{2-} ion. As a result, \text{Ca}^{2+} and {\text{SO}_4}^{2-} ions in calcium sulfate exist on a 1:1 ratio.

2\;\text{LiOH} +\text{H}_2\text{SO}_4 \to \text{Li}_2\text{SO}_4 + 2\;\text{H}_2\text{O}.

Ammonia, NH₃

Ammonia NH₃ can also act as a base and neutralize acids. NH₃ exists as NH₄OH in water:

\text{NH}_3 + \text{H}_2\text{O} \to \textbf{NH}_{\bf 4}\text{OH}.

The ion {\text{NH}_4}^{+} acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt:

2\;\textbf{NH}_{\bf 4}\text{OH}+ \text{H}_2\text{SO}_4 \to (\textbf{NH}_{\bf 4})_2\text{SO}_4+2\;\text{H}_2\text{O}.

The formula of the salt (NH₄)₂SO₄ in the fourth option spelled the ammonium ion incorrectly.

As part of the salt (NH₄)₂SO₄, the ammonium ion NH₄⁺ is one of the products of this reaction and can't neutralize H₂SO₄ any further.

7 0
2 years ago
PLEASE HELP I WILL GIVE YOU BRAINLIEST Provide the chemical formula for the following
a_sh-v [17]
1. CaCr2O7
2. Sodium sulfite
3. (NH4)2SO3
4. Copper (II) nitrite
8 0
3 years ago
In the presence of excess oxygen, methane gas burns in a constant-pressure system to yield carbon dioxide and water: CH4 (g) 2O2
Degger [83]

Answer:

The energy released will be -94.56 kJ or -94.6 kJ.

Explanation:

The molar mass of methane is 16g/mol

The given reaction is:

CH_{4}(g) + 2O_{2} (g) --> CO_{2} (g)+ 2H_{2}O(l)

the enthalpy of reaction is given as  ΔH = -890.0 kJ

This means that when one mole of methane undergoes combustion it gives this much of energy.

Now as given that the amount of methane combusted = 1.70g

The energy released will be:

=\frac{energy released by one moleXgiven mass}{molarmass} =\frac{-890X1.7}{16}= -94.56 kJ

7 0
3 years ago
Chemist A burns 10.000 grams of C₂H₆ in a constant-volume container at 25ºC. Chemist B burns an unknown mass of C₂H₆ in a consta
alexandr1967 [171]

Answer:

10.000 grams

Explanation:

For the first law of thermodynamics, the energy must be conserved, that means that the energy in form of heat (Q) must be equal to the sum of work (W) and internal energy(ΔU) :

Q = W + ΔU

ΔU depends on the temperature and W in the variation of pressure and volume. Q depends on the temperature, but also the mass. So, there is the same temperature, ΔU is equal for both reaction, if there is no work done, the heat must be equal for both of them. So the mass such be the same.

5 0
3 years ago
A biochemist carefully measures the molarity of magnesium ion in of cell growth medium to be . Unfortunately, a careless graduat
denis23 [38]

Answer:

760 uM

Explanation:

<em>A biochemist carefully measures the molarity of magnesium ion in 47, mL of cell growth medium to be 97 uM. Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 6.0 mL. Calculate the new molarity of magnesium ion in the cell growth medium Be sure your answer has the correct number of significant digits.</em>

The problem here is that the amount of magnesium ion remains the same irrespective of the volume.

Amount of magnesium in the growth medium = <em>molarity x volume</em>

    = 97 x 10^{-6} x 47 x 10^{-3} = 4.559 x 10^{-6}

Then, the volume reduced to 6.0 mL, the new molarity becomes;

<em>molarity = mole/volume </em>

       = 4.559 x 10^{-6}/6 x 10^{-3} = 7.598333 x 10^{-4} M = 759.83333 uM

To the correct number of significant digits = 760 uM

6 0
2 years ago
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