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2 years ago

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Chemistry

2 answers:

grigory [225]2 years ago

6 0

Extrusive. Please give me a brainliest! Im new i really need it thanks

Svetradugi [14.3K]2 years ago

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Your answer for that question is D

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The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

kherson [118]

Answer: The molecular formula of the compound is $C_4H_{10}O_4$

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

Calculating the molecular formula:

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=36.26g$

Mass of $H_2O=14.85g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

4 0

3 years ago

How many milligrams of sodium sulfide are needed to completely react with 25.00 ml of a 0.0100 m aqueous solution of cadmium nit

NARA [144]

Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)

v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol

n(Na₂S)=n{Cd(NO₃)₂}=cv

m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv

m(Na₂S)=78.046*0.0100*25.00≈19.5 mg

5 0

3 years ago

Read 2 more answers

¿Cuáles son las formas alotrópicas del carbono (como se encuentra en la naturaleza)?

Mars2501 [29]

Answer: El carbono, que en estado sólido, puede adoptar muchas formas alotrópicas, siendo las más comunes el diamante (red tridimensional) y el grafito (láminas), aunque también puede formar nanoestructuras en forma de balón de fútbol (fullerenos) o tubos diminutos (nanotubos de carbono), entre otras posibilidades.

Explanation:

3 0

3 years ago

Is matter lost when a candle is burned

zlopas [31]

No it is redistributed and the state changes to gas and liquid

5 0

3 years ago

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PLEASE HELP ASAP WILL GIVE BRAINLIEST

kirill115 [55]

The amount of heat required to convert H₂O to steam is : 382.62 kJ

Given data :

Mass of liquid water ( m ) = 150 g

Temperature of liquid water = 43.5°C

Temperature of steam = 130°C

<h3 /><h3>Determine the amount of heat required </h3>

The amount of heat required = ∑ q1 + q2 + q3 ----- ( 1 )

where ;

q1 = heat required to change Temperature of water from 43.5°C to 100°C . q2 = heat required to change liquid water at 100°C to steam at 100°C

q3 = heat required to change temperature of steam at 100°C to 130°C

For q1

M* S $_{water}$ *ΔT

= 150 * 4.18 * ( 100 - 43.5 )

= 35425.5 J

For q2

moles * ΔHvap

= (150 / 18 )* 40.67 * 1000

= 338916.67 J

For q3

M * S $_{steam}$ * ΔT

= 150 * 1.84 * ( 130 -100 )

= 8280 J

Back to equation ( 1 )

Amount of heat required = 35425.5 + 338916.67 + 8280 = 382622.17 J

≈ 382.62 kJ

Hence we can conclude that The amount of heat required to convert H₂O to steam is : 382.62 kJ.

Learn more about Specific heat of water : brainly.com/question/16559442

6 0

2 years ago