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patriot [66]
3 years ago
14

Rock salt is a mixture of salt (sodium chloride) and sand. Describe how you would separate rock salt to obtain salt crystals and

pure dry sand. In this question you get marks for how well your answer is written. You will get marks for: spelling, grammar, organising your ideas and information clearly and using key scientific words.
Chemistry
1 answer:
choli [55]3 years ago
6 0

Answer:

pour the rock salt mixture throught a filter made from paper and allow the liquid to filtrate .

Explanation:

Separating Sand and Salt

Probably the easiest method to separate the two substances is to dissolve salt in water, pour the liquid away from the sand, and then evaporate the water to recover the salt.

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25. What is the equilibrium constant expression for the following? 2Hg (g) + O₂(g) → 2HgO (s) A. K= [Hg] [0₂]. B. K= [HgO]/([Hg]
boyakko [2]

Answer:

2 Hg (g) + o2 ---> 2 H2o

equilibrium constant

K = (c) (d) / (a)(b)

K= ( H2o)²/(He) ²(o2)

5 0
2 years ago
Do covalent bonds dissolve in methanol
ArbitrLikvidat [17]
Yes,covalent bonds do dissolve in methanol


5 0
4 years ago
Given that the molar mass of NaCl is 58.44 g/mol, Solve for the molarity of a solution that contains 87.75 g of NaCl in 500 mL o
aliina [53]

Molarity is defined as the number of moles of solute per 1 L of solvent.

the mass of NaCl in the solution is 87.75 g

number of moles of NaCl is calculated by dividing mass present by molar mass

number of NaCl moles = 87.75 g / 58.44 g/mol = 1.502 mol

the number of NaCl moles in 500 mL is - 1.502 mol

therefore number of NaCl moles in 1000 mL is - 1.502 mol/ 500 mL x 1000 mL/L = 3.004 mol

molarity of NaCl is - 3.004 M

answer is D. 3.00 M

5 0
3 years ago
Read 2 more answers
Sunflower oil contains 0.080 mol palmitic acid (C16H32O2)/mol, 0.060 mol stearic acid (C18H36O2)/mol, 0.27 mol oleic acid (C18H3
Cerrena [4.2K]

Answer:

x_H=0.882

x_N=0.118

Explanation:

In this reactor, oleic and linoleic acid react with hydrogen to form stearic acid. This reactions can be represented by:

Oleic: C_{18}H_{34}O_2 (l) + H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)

Linoleic: C_{18}H_{32}O_2 (l) + 2 H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)

Having this reactions in mind, the first thing is to determine the moles of hydrogen required:

<u>Base of caculation: 1 mol of sunflower oil</u>

For oleic acid: n_{Holeic}=\frac{1 mol H_2}{1 mol oleic}*\frac{0.27 mol oleic}{1 mol oil}*\frac{335 mol oil}{hr}

n_{Holeic}=frac{90.45 mol H_2}{hr}

For linoleic acid: n_{Hlinoleic}=\frac{2 mol H_2}{1 mol linoleic}*\frac{0.59 mol linoleic}{1 mol oil}*\frac{335 mol oil}{hr}

n_{Holeic}=frac{395.3 mol H_2}{hr}

n_{Htotal}=\frac{90.45 mol H_2}{hr}+\frac{395.3 mol H_2}{hr}

n_{Htotal}=frac{485.75 mol H_2}{hr}

Applying the excess:

n_{Htotal}=frac{485.75 mol H_2}{hr}*1.65=801.48 mol

Nitrogen: n_N= 801.48 mol*\frac{0.05 mol N}{0.95 mol}

n_N= 42.2 mol N

<u>After the reactions</u>:

n_H=801.48 mol-485.75mol=315.73 mol

and the nitrogen is inert.

Purge stream:

n_total=42.2+315.73 mol=357.93 mol

x_H=\frac{315.73mol}{357.93mol}=0.882

x_N=\frac{42.2mol}{357.93mol}=0.118

4 0
3 years ago
Hi I need help with the Lab report for the 'Flame Lab' in edge
Ludmilka [50]

Flame test is a qualitative test which is used to identify the metal and metalloid ion in the sample.

<h3>What is Flame Test ?</h3>

A Flame test is used to identify the metal and metalloid ion in the sample. Flame test is a qualitative test. Not every metal ion emit color when it is heated in the gas burner.

<h3>What is the Purpose of Flame test ?</h3>

The purpose of flame test is used to find the identities of ions in two solutions of unknown composition by comparing the colors they produce.

<h3>What are the material used ?</h3>
  • Bunsen Burner
  • Matches
  • Gloves
  • Weighing dishes
  • Beakers

Thus from the above conclusion we can say that Flame test is a qualitative test which is used to identify the metal and metalloid ion in the sample.

Learn more about the Flame test here: brainly.com/question/864891
#SPJ1

7 0
2 years ago
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