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notsponge [240]
3 years ago
15

How many mol of Cl are in 10.0 g of Cl? The molar mass of Cl is 35.45 g/mol

Chemistry
2 answers:
Delvig [45]3 years ago
6 0

Answer:

\boxed {\boxed {\sf 0.282 \ mol \ Cl}}

Explanation:

The molar mass, or mass of 1 mole of a substance, is used to convert grams to moles. We are given the molar mass of chlorine. It is 35.45 grams per mole.

We convert using dimensional analysis, so we must set up a conversion factor using the molar mass.

\frac { 35.45 \ g \ Cl}{1 \ mol \ Cl}

We are converting 10.0 grams of chlorine to moles, so we must multiply the conversion factor by this value.

10.0 \ g  \ Cl* \frac { 35.45 \ g \ Cl}{1 \ mol \ Cl}

Flip the conversion factor. The result is equivalent and allows us to cancel the units of grams of chlorine.

10.0 \ g  \ Cl* \frac {1 \ mol \ Cl}{ 35.45 \ g \ Cl}

10.0 * \frac {1 \ mol \ Cl}{ 35.45 }

\frac {10.0 }{35.45} \ mol \ Cl

0.282087447 \ mol \ Cl

The original value of grams (10.0) has 3 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 0 in the ten-thousandth place tells us to leave the 2 in the thousandth place.

0.282 \ mol \ Cl

Approximately <u>0.282 moles of chlorine</u> are in 10.0 grams of chlorine.

postnew [5]3 years ago
5 0
There are 3.545 mol in 10 g of CL
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Quinine (C20H24N2O2) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine
dolphi86 [110]

Answer:

The pH of saturated solution of the quinine is 10.05

Explanation:

Quinine (Q) is C20H24N2O2 has a molar mass of 324.4 g/mol

Q can behave as a weak base. Kb and pKb can be calculated for weak bases

pKb1 is to be considered when solving the question.

pKb1 = 5.1

Step 1 : Calculate the Kb of Quinine

            pKb1 = - log [kb]

                5.1  = - log [kb]

                take Antilog of both side

             [kb] = 7.94 x 10∧-6

Step 2: Calculate the concentration of saturated solution of Q in mol/dm3

           From the question, 1900 ml of solution contains 1 g of Q

           Therefore,  1000 ml of solution will contain........... x g of Q

           x = 1000 /1900

           x = 0.526 g in 1 dm3

In calculating concentration in mol/dm3,

Concentration in mol/dm3 = concentration in g /dm3 divided by molar mass

Molar mass of Q = 324.4

Concentration in mol/dm = 0.526 /324.4

                                         = 0.0016 mol/dm3

Step 3: Calculating the Concentration of OH-

            At Equilibrium, Kb = x² / 0.0016

            7.94 x 10∧-6 = x² / 0.0016

            x = √ 0.0016 × 7.94 x 10∧-6

            x = 1. 127 × 10∧-4 mol/dm3

The concentration of OH- = 1. 127 × 10∧-4 mol/dm

Step 4:  Calculating the pH of Quinine

           Recall, pOH = - log [OH-]

           pOH = - log [1. 127 × 10∧-4]

           pOH = 3.948

           Also recall that pH + pOH = 14

           pH = 14 - 3.948

           pH = 10.05

           

8 0
4 years ago
1. Which of the following equations is balanced?
pantera1 [17]
The answer would be A
6 0
3 years ago
Look at the reaction below.
HACTEHA [7]
The acid is H2SO4(an)
6 0
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From the following reaction and data, find (a) S o of SOCl2 (b) T at which the reaction becomes nonspontaneous SO3(g) + SCl2(l)
disa [49]

Answer:

618 J/Kmol

T > 1.36 x 10³ K

Explanation:

The  balanced reaction of interest is:

                           SO₃ (g) + SCl₂ (l) ⇒    SOCl₂ (l) +     SO₂ (g)

with the data:

ΔHºf (kJ/mol)      -396          -50.0          -245.6         -296.8

Sº(J/mol·K)             256.7       184               ?               -248.1

ΔGº=  -75.2 kJ

We know, we can find the standard  change inGibb´s free energy from the equation:

ΔGºrxn =  ΔHºrxn - TΔSºrxn

So we can calculate ΔHºrxn  = ∑ ΔHºf prod  -  ΔHºreact, and substitute into this equation to solve Sº SOCl₂.

ΔHºrxn = ( -245.6 + (-296.8) ) - ( -396 - 50) kJ = - 96.4 kJ

Similarly  for ΔSºrxn

 ΔSºrxn = (-0.248.1 +Sº SOCl₂) - (0.256.7 +0.184) kJ/K

= -0.689 kJ /K -+ Sº SOCl₂

Plugging the values for the expression for  ΔGºrxn:

-75.2 kJ = -96.4 kJ - 298 K  x  ( -0.689 kJ /K + Sº SCl₂ )

-75.2 kJ = -96.4 kJ + 205.3 Kj - 298 Sº SCl₂

-184  kJ = -298 K  x Sº SCl₂

0.618 kJ/molK = Sº SCl₂

= 0.618 kJ/K x 1000 J = 618 J/Kmol

For the second part we will still be using the Gibb´s free energy change  equation as above , but now we will solve for T when the reaction becomes  non-spontaneous.

For the reaction to become non-spontaneous  ΔGº is positive, and this happens when the term  TΔSº becomes greater tha ΔHº:

ΔGºrxn =  ΔHºrxn - TΔSºrxn

0 =   ΔHºrxn - TΔSºrxn ⇒  TΔSºrxn  =  ΔHºrxn

                                           T= ΔHºrxn / ΔSºrxn

ΔSºrxn  = -0.689 J/Kmol + 0.618 J/Kmol = -0.0710 kJ/Kmol

( using the value  the value just calculated from above )

T =  - 96.4 kJ / -0.071 kJ/K = 1.36 x 10³ K

For temperatures greater than 1.36 x 10³ K the reaction becomes non-spontaneous.

4 0
4 years ago
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