Which method of drawing hydrocarbons is the fastest to draw?

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl),

skad [1K]

The following are the answers to the different questions:
<span>The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?

A. 101.53° C

Which of the following solutions will have the lowest freezing point?

D. 1.0 mol/kg magnesium fluoride (MgF2)

Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?

A. sodium iodide (NaI)

Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?

B. 96.3 kPa

Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?

D. 108.1°C</span>

4 0

2 years ago

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An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered

Karo-lina-s [1.5K]

Answer:

Pb(NO3)2

Cd(NO3)2

Na2SO4

Explanation:

In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.

When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.

After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.

The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.

5 0

1 year ago

A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t

TEA [102]

Explanation:

The reaction equation will be as follows.

$Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)$

So, according to this equation, 1 mole $Ca(OH)_{2}$ = 2 mol HBr = 1 mol $CaBr_{2}$

Therefore, calculate the number of moles of calcium hydroxide as follows.

No. of moles of $Ca(OH)_{2}$ = $V \times Molarity$

= $50 \times 0.6$

= 30 mmol

Similarly, calculate the number of moles of HBr as follows.

No. of moles of HBr = $M \times V$

= $50 \times 0.6$

= 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of $CaBr_{2}$ = $30 \times \frac{1}{2}$

= 15 mmol

Hence, calculate the amount of heat released as follows.

Heat released in the reaction(q) = $m \times s \times \Delta T$

as, m = mass of solution

and, Density = $\frac{mass}{volume}$

or, mass = Density × Volume

= 1.08 g/ml \times (50 + 50) ml

= 108 g

where, s = specific heat of solution = 4.18 j/g.k

and, change in temperature $\Delta T$ = $(26 - 23)^{o}C$

= $3
^{o}C$

Hence, the heat released will be as follows.

q = $m \times s \times \Delta T$

q = $108 \times 4.18 \times 3^{o}C$

= 1354.32 joule

or, = 1.354 kJ (as 1 kJ = 1000 J)

Also, $\Delta H_{rxn}$ = $\frac{-q}{n}$

= $\frac{-1.354}{15 \times 10^{-3}}$

= -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

6 0

2 years ago

Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f

Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0

2 years ago

The general term for a large molecule made up of many similar subunits is

Setler79 [48]

A) Polymer is the general name of large units made of many smaller units (these would be called monomers). An example is starch, this is a carbohydrate polymer that is made up of smaller units (monomers) called glucose.

6 0

2 years ago