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sweet-ann [11.9K]
2 years ago
7

Determine the method to separate a mixture of ground glass and salt in order to have a pile of ground glass and another pile of

salt.
write a step by step procedure that would accomplish this task. note any safety concerns in your procedures
Chemistry
1 answer:
Serga [27]2 years ago
3 0
1 Water is added to the mixture of ground glass and salt to dissolve the salt content

2 The mixture is then filtered leaving the grounded glass particles in the sieve

3 The salt is then evaporated leaving the salt crystals
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The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product str
iren2701 [21]

Answer:

Explanation:

The first thing to do is to write out the chemical equation showing the reaction between the two chemical compounds.

C2H4 + HBr ------------------------> C2H5Br.

The chemical species/compounds reacted in the ratio of 1 : 1 : 1. That is one mole of C2H4 reacted with one mole of HBr to give one mole of C2H5Br.

So, we are given; C2H5Br = 58.7 mol%, HBr = 18.3 mol%. Thus, for C2H4, we have [ 100 - (58.7 + 18.3)] = 23 mol%. Therefore, the limiting reagent is HBr.

So, if we have 100 mol/s ;

C2H5Br = 100 × 0.587 = 58.7 mol/s, HBr = 100 × 0.183 = 18.3 mol/s and C2H4 = 100 × 0.23 = 23 mol/s.

Therefore, the total value of the flow rate in the feed = (HBr in feed) mol/s + (C2H4 in feed) mol/s = (58.7 + 18.3)mol/s + (58.7 + 23) mol/s =( 77 + 81.7) mol/s = 158.7 mol/s. .

So, 100/158.7 × 30.5 = 192.2.

Also the feed composition in HBr and C2H4 is;

HBr = 77/158.7 = 0.4852, C2H4 = 81.7/158.7 = 0.515.

Therefore, the number of HBr in feed and C2H4 in feed = 160m/s and 200m/s respectively (approx.)

Therefore, the fraction conversion is;

8 0
2 years ago
A sample compound contains 9.11 g Ni and 5.89g F. What is the empirical formula of this compound?
Alexxx [7]

The empirical formula of the compound is C. NiF₂.

<em>Step 1</em>. Calculate the <em>moles of each element</em>

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of Ni to F.

Moles of Ni = 9.11 g Ni × (1 mol Ni /(58.69 g Ni) = 0.1552 mol Ni

Moles of F = 5.89 g F × (1 mol F/19.00 g F) = 0.3100 mol F

<em>Step 2</em>. Calculate the <em>molar ratio</em> of the elements

Divide each number by the smallest number of moles

Ni:F = 0.1552:0.3100 = 1:1.997 ≈ 1:2

<em>Step 3</em>: Write the <em>empirical formula</em>

EF = NiF₂

8 0
3 years ago
Read 2 more answers
How many atoms of cobalt are in one formula unit of cobalt (III) oxide? one two three six
IrinaK [193]
I believe the correct answer is the second option. There will be two cobalt atoms in one formula unit of cobalt (III) oxide. It has a chemical formula of Co2O3. This compound is does not naturally occur so it is being synthesized. It is mostly used as bleaching agent.
7 0
3 years ago
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Based on the kinetic theory, which statement is true? (5 points)
Hoochie [10]

Answer:

the particles of matter are arranged in different ways for the different states

Explanation:

because solid liquid and gas all three matters have different states for example the particles in a solid are closely packEd and form of movement is vibration

7 0
2 years ago
If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

4 0
3 years ago
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