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Misha Larkins [42]
2 years ago
15

What happens as a result of the kinetic energy of the particles in a liquid being comparable with the intermolecular attractions

between the particles?
A The particles do not move enough to change positions, so the liquid is not compressible.
B The particles move enough that they separate from one another and take the shape of their container.
C The particles do not move enough, so they are fixed in place, and the liquid has a fixed volume.
D The particles move enough that they are not fixed in place, and the liquid can flow.
Chemistry
2 answers:
sdas [7]2 years ago
6 0

Answer : Option D) The particles move enough that they are not fixed in place, and the liquid can flow.

Explanation : The kinetic energy of the particles are allowed to move freely and are in motion when in the liquid state whereas the intermolecular particles can just flow; as the intermolecular attractions between the particles allows the liquid to flow by giving them a force to flow.

marusya05 [52]2 years ago
4 0

Answer:

Option D) The particles move enough that they are not fixed in place, and the liquid can flow.

Explanation:

Took the test just now

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The density of a gaseous compound is 1.623 g/l at stp determine the molar mass of the compound
victus00 [196]
<span>The molar mass of the compound is 36.355 g/mol. This is calculated by knowing that 1 mol of gas fills 22.4 L of volume, so 1.623 g/L = X g/mol * 1/22.4 mol/L -> 1.623 g/L * 22.4 L/mol = X g/mol -> 36.355 g/mol = X g/mol</span>
4 0
2 years ago
Salt in crude oil must be removed before the oil undergoes processing in a refinery. The
irina1246 [14]

Answer:

\large \boxed{0.64 \, \%}

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}

2. Calculate the mass of salt in the salty oil

\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}

3. Calculate the mass of salt in the spent water

\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g  

5. Concentration of salt in washed oil

\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}

3 0
2 years ago
2. If 2.50g of sodium hydroxide is being reacted with 4.30g of magnesium chloride, how many grams of magnesium hydroxide would b
Virty [35]

Answer:

1.822 g of magnesium hydroxide would be produced.

Explanation:

Balanced reaction: 2NaOH+MgCl_{2}\rightarrow Mg(OH)_{2}+2NaCl

     Compound                                 Molar mass (g/mol)

         NaOH                                           39.997

         MgCl_{2}                                           95.211

        Mg(OH)_{2}                                        58.3197

So, 2.50 g of NaOH = \frac{2.50}{39.997} mol of NaOH = 0.0625 mol of NaOH

      4.30 g of MgCl_{2}  = \frac{4.30}{95.211} mol of MgCl_{2} = 0.0452 mol of MgCl_{2}

According to balanced equation-

2 mol of NaOH produce 1 mol of Mg(OH)_{2}    

So, 0.0625 mol of NaOH produce (\frac{0.0625}{2}) mol of NaOH or 0.03125 mol of NaOH

1 mol of MgCl_{2} produces 1 mol of Mg(OH)_{2}

So, 0.0452 mol of MgCl_{2} produce 0.0452 mol of Mg(OH)_{2}

As least number of moles of Mg(OH)_{2} are produced from NaOH therefore NaOH is the limiting reagent.

So, amount of Mg(OH)_{2} would be produced = 0.03125 mol

                                                                           = (0.03125\times 58.3197) g

                                                                           = 1.822 g

6 0
2 years ago
Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (Na
Afina-wow [57]

The pH of a buffer solution : 4.3

<h3>Further explanation</h3>

Given

0.2 mole HCNO

0.8 mole NaCNO

1 L solution

Required

pH buffer

Solution

Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.

\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}

valence according to the amount of salt anion  

Input the value :

\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3

7 0
2 years ago
4. Convert the following: a. 4g mol of MgCl2 to g b. 2 lb mol of C3H8 to g c. 16 g of N2 to lb mol d. 3 lb of C2H6O to g mol
Nuetrik [128]

Answer:

a) 381.2 g

b) 39916 g

c) 0.0013 lb mol

d) 29.6 g mol

Explanation:

The molecular weight (mw) of a compound is the mass of it per mole, so it's the ratio of the mass (m) per mole (n).

a) The molecular weight of one mol is found at the periodic table. So, for Mg, mw = 24.3 g/mol, for Cl = 35.5 g/mol, so for MgCl2, mw = 24.3 + 2*35.5 = 95.3 g/mol. The g mol is the mass divided by the molecular weight:

g mol = m/mw

4 = m/95.3

m = 381.2 g

b) The pound (lb) is a unity of mass, and the lb mol is a unity of the mass divided by the molecular weight. So, by the periodic table, the molecular weight of C3H8 is 3*12 (of C) + 8*1 (of H) = 44 lb/mol.

lb mol = m/mw

2 = m/44

m = 88 lb

1 lb = 453.592 g

So, m = 88*453.592 = 39916 g

c) The molecular weight of N2 is 2*14 (of N) = 28 lb/mol.

m = 16/453.592 = 0.0353 lb

lb mol = m/mw

lb mol = 0.0353/28

lb mol = 0.0013 lb mol

d) The molecular weight is 2*12 (of C) + 6*1(of H) + 1*16(of O) = 46 g/mol

3 lb = 1360.78 g

g mol = m/mw

g mol = 1360.78/46

g mol = 29.6 g mol

6 0
2 years ago
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