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3 years ago

I need help with these two questions

Chemistry

1 answer:

natulia [17]3 years ago

8 0

I'm not sure about 2 but in 3 it would float

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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

The elements in a row of the the periodic table

Aleks [24]

Answer:

Rows in the periodic table are called periods. As one moves from left to right in a given period, the chemical properties of the elements slowly change. Columns in the periodic table are called groups. Elements in a given group in the periodic table share many similar chemical and physical properties

Explanation:

5 0

2 years ago

Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?

chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now we start with 150 grams of Thorium-234

$150 \times \frac{1}{2}=24.1$

$75 \times \frac{1}{2} =48.2$

$34.5 \times \frac{1}{2} =72.3$

$17.25 \times \frac{1}{2} =96.4$

$8.625\times \frac{1}{2} =120.5$

So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

$A=A_{0} \cdot \frac{1}{2^{n}}$

Where

$A_0$ is the initial sample amount

n = the number of half-lives that pass in a given period of time.

7 0

3 years ago

Which statements are true of all scientific endeavors? Check all that apply.

dybincka [34]

Scientific endeavors is basically all the things that would contribute to the process of achieving a certain scientific knowledge, from the initial observation, up to the point until we can hold that certain thing as acknowledged truth. So,

All scientific endeavors are supported by evidence.
All scientific endeavors are a systemic process.
All scientific endeavors involve observation.
All scientific endeavors involve experimentation.
All scientific endeavors involve the collection of information.

4 0

3 years ago

Free-energy change, ΔG∘, is related to cell potential, E∘, by the equation ΔG∘=−nFE∘ where n is the number of moles of electrons

elena-s [515]

Answer:

-372000 J or -372 KJ

Explanation:

We have the electrochemical reaction as;

Mg(s) + Fe^2+(aq)→ Mg^2+(aq) + Fe(s)

We must first calculate the E∘cell from;

E∘cathode - E∘anode

E∘cathode = -0.44 V

E∘anode = -2.37 V

Hence;

E∘cell = -0.44 V -(-2.37 V)

E∘cell = 1.93 V

n= 2 since two electrons were transferred

F=96,500C/(mol e−)

ΔG∘=−nFE∘

ΔG∘= -( 2 * 96,500 * 1.93)

ΔG∘= -372000 J or -372 KJ

4 0

3 years ago