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4 years ago

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Chemistry

1 answer:

zzz [600]4 years ago

4 0

Answer:

soorry i ddsmfcj know

Explanation:

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A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

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

Formula used :

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

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur

lbvjy [14]

Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 40.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g$

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

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

q = heat absorbed by water

$m$ = mass of water = 40.0 g

$T_{final}$ = final temperature of water = 20.0°C

$T_{initial$ = initial temperature of water = 17.0°C

$c$ = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

$q=40.0\times 4.186\times (20.0-17.0)]$

$q=502J$

Hence, the joules of heat were re-leased by the lead is 502

5 0

3 years ago

Which would be an example of a compound?<br> oxygen<br> carbon dioxide<br> sand<br> carbon

GenaCL600 [577]

Answer:

carbondioxide is an example of compound.

4 0

4 years ago

The dissociation of a/an ________ releases hydrogen ions and increases the concentration of hydrogen ions in a solution.

Crank

Acid...hope it helps

5 0

3 years ago

How old is a bone if it now .3125 of C-14 when it originally had 80.0g of C-14

Taya2010 [7]

<h3>Answer:</h3>

42960 years

<h3>Explanation:</h3>

<u>We are given;</u>

Remaining mass of C-14 in a bone is 0.3125 g
Original mass of C-14 on the bone is 80.0 g
Half life of C-14 is 5370 years

We are required to determine the age of the bone;

Using the formula;

Remaining mass = Original mass × 0.5^n , where n is the number of half lives.

Therefore;

0.3125 g = 80.0 g × 0.5^n

3.90625 × 10^-3 = 0.5^n

Introducing logarithm on both sides;

log 3.90625 × 10^-3 = n log 0.5

Solving for n

n = log 3.90625 × 10^-3 ÷ log 0.5

= 8

Therefore, the number of half lives is 8
But, 1 half life is 5370 years
Therefore;

Age of the rock = 5370 years × 8

= 42960 years

Thus, the bone is 42960 years old

7 0

3 years ago