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Juli2301 [7.4K]

2 years ago

10

a compound had a molecular weight of 112.124 atomic mass units and the empirical formula C3H4O what is the molecular formula of

the compound?

1 answer:

Snowcat [4.5K]2 years ago

8 0

Answer : The molecular of the compound is, $C_6H_8O_2$

Solution : Given,

Molecular weight of compound = 112.124 amu

The Empirical formula = $C_3H_4O_1$

First we have to calculate the empirical formula weight.

The empirical formula weight = 12(3) + 4(1) + 1(16) = 56 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}\\\\n=\frac{112.124}{56}=2$

Molecular formula = $(C_3H_4O_1)_n=(C_3H_4O_1)_2=C_6H_8O_2$

Therefore, the molecular of the compound is, $C_6H_8O_2$

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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

Ka, the acid dissociation constant, for an acid is 9 × 10^{−4} at room temperature. At this temperature, what is the approximate

devlian [24]

Dissociation=k×no of moles
percentage of dissociation=9.0×10^-4×1×100
knowing that x%=x/100,we then say;
x/100=9.0×10^-4×1×100
therefore, x=100×100×9×10^-4×1
x=9
x percentage of dissociation=9%

8 0

2 years ago

Using complete subshell notation (not abbreviations, 1s 22s 22p 6 , and so forth), predict the electron configuration of each of

dybincka [34]

<u>Answer:</u> The electronic configuration of the elements are written below.

<u>Explanation:</u>

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

For the given options:

<u>Option a:</u> Carbon (C)

Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.

The electronic configuration of carbon is $1s^22s^22p^2$

<u>Option b:</u> Phosphorus (P)

Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.

The electronic configuration of phosphorus is $1s^22s^22p^63s^23p^3$

<u>Option c:</u> Vanadium (V)

Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.

The electronic configuration of vanadium is $1s^22s^22p^63s^23p^64s^23d^3$

<u>Option d:</u> Antimony (Sb)

Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.

The electronic configuration of antimony is $1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3$

<u>Option e:</u> Samarium (Sm)

Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.

The electronic configuration of samarium is $1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^6$

Hence, the electronic configuration of the elements are written above.

4 0

2 years ago

What is the density of a piece of metal in g/cm3 if its mass is determined to be 42.20 g and it is in the shape of a cube, with

maria [59]

It's a cube so the volume = edge^3
Volume = 2.5^3 cm^3 = 15.625 cm^3

density = mass / volume = 42.20 / 15.625 = 2.70 You have 3 places of accuracy.

density of object = 2.70 grams / cm^3 <<<<=== answer.

4 0

2 years ago

Read 2 more answers

How much heat is lost when changing 65 g of water vapor (H2O) at 421 K to ice at 139 K?

poizon [28]

The heat change will be

Moles of water = mass / Molar mass = 65/ 18 = 3.61 mol

specific heat of ice =2.09J /g C

specific heat of water = 4.184 J/g C

Specific heat of vapour= 2.01 /g C

Heat of fusion = 3.33X10⁵ J /kg = 333 J /g

Heat of vaporization = 2.26 X10⁶J/kg = 2260J/g

Q1 = heat change when vapours get cooled to 373.15 K

Q2 = heat change when vapours get converted to liquid water

Q3 = heat change when liquid water cools to 273.15 K

Q4= heat change when liquid water freezes to ice

Q5= heat change when ice cools from 273.15K to 139 K

Q1= mass of water X specific heat of vapours X change in temperature

Q1 = 65 X 2.01 /g C X (421-373.15) = 6251.60 J = 6.252 kJ

Q2 = heat of vaporization X mass = 2260 X 65 = 146900 = 146.9 kJ

Q3 = mass X specific heat of water X change in temperature =

Q3 = 65 X 4.184 X (373.15-273.15) = 65 X 4.184 X 100 = 27196 J = 27.196kJ

Q4 = heat of fusion X mass =333X65 = 21645 J = 21.645 kJ

Q5 = mass X specific heat of ice X change in temperature

Q5 = 65 X 2.09 X (273.15-139) = 18224.3 J = 18.224 kJ

Total energy = 6.252 +146.9+27.196+ 21.645+ 18.224 = 220.217

As this is energy released so it will be expressed in negative

-220.217

from the given options the correct answer will be -219.4 kJ

The answer is little different as the reference values of specific heats or enthalpy may vary.

3 0

2 years ago