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02.07.2021

What is the molarity of the following solution? 3.48 mol in 4.1 L of solution

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24.06.2023, solved by verified expert

Ranjana Bhatt

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molarity is moles/L so take the 3.48 and divide it by 4.1L

and get 0.848... and need to keep your answer to 2 digits as that is th emost precise measurement you have (the 4.1L)

so 0.85mol/L

Explanation:

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Chemistry

What is the molarity of the following solution? 3.48 mol in 4.1 L of solution

Step-by-step answer

P Answered by PhD

molarity is moles/L so take the 3.48 and divide it by 4.1L

and get 0.848... and need to keep your answer to 2 digits as that is th emost precise measurement you have (the 4.1L)

so 0.85mol/L

Explanation:

Chemistry

How many atoms are present in a 1.01g sample of Phosphorus (P)? [Note: 1 mol = 6.022x10²³ atoms, molar mass of Phosphorus =30.97 g/mol) * 1 point Captionless Image 3.48 x 10²¹ atoms 1.97 x 10²² atoms 1.04 x 10²⁶ atoms 3.36 x 10²⁴ atoms

Step-by-step answer

P Answered by PhD

$\boxed {\boxed {\sf B. \ 1.97*10^{22} \ atoms }}$

Explanation:

First, convert the grams to moles, then convert moles to atoms.

1. Grams to moles

To convert from grams to moles, we use the molar mass. Phosphorus's molar mass is given to use.

Phosphorus (P)= 30.97 g/mol

We can use this number as a fraction.

$\frac{30.97 \ g \ P }{ 1 \ mol \ P }$

Multiply by the given number of grams (1.01)

$1.01 \ g \ P *\frac{30.97 \ g \ P }{ 1 \ mol \ P }$

Flip the fraction so the grams of phosphorus cancel out.

$1.01 \ g \ P *\frac{1 \ mol \ P }{ 30.97 \ g \ P}$

$1.01 *\frac{1 \ mol \ P }{ 30.97 }$

$\frac{1.01 \ mol \ P }{ 30.97 }$

$0.03261220536 \ mol \ P$

2. Convert Moles to Atoms

To convert from moles to atoms, we use Avogadro's Number.

6.022*10²³

This tells us the number of particles in 1 mole. In this case, it is atoms of phosphorus in 1 mole of phosphorus. We can create a fraction.

$\frac{6.022 *10^{23} \ atoms \ P }{1 \ mol \ P}$

Multiply by the number of moles we calculated.

$0.03261220536 \ mol \ P*\frac{6.022 *10^{23} \ atoms \ P }{1 \ mol \ P}$

The moles of phosporus will cancel.

$0.03261220536 *\frac{6.022 *10^{23} \ atoms \ P }{1 }$

$0.03261220536 *{6.022 *10^{23} \ atoms \ P }$

$1.96390701*10^{22} \ atoms \ P$

This is closest to 1.97*10²² atoms, so choice B is correct.

Chemistry

Step-by-step answer

P Answered by PhD

$\boxed {\boxed {\sf B. \ 1.97*10^{22} \ atoms }}$

Explanation:

First, convert the grams to moles, then convert moles to atoms.

1. Grams to moles

To convert from grams to moles, we use the molar mass. Phosphorus's molar mass is given to use.

Phosphorus (P)= 30.97 g/mol

We can use this number as a fraction.

$\frac{30.97 \ g \ P }{ 1 \ mol \ P }$

Multiply by the given number of grams (1.01)

$1.01 \ g \ P *\frac{30.97 \ g \ P }{ 1 \ mol \ P }$

Flip the fraction so the grams of phosphorus cancel out.

$1.01 \ g \ P *\frac{1 \ mol \ P }{ 30.97 \ g \ P}$

$1.01 *\frac{1 \ mol \ P }{ 30.97 }$

$\frac{1.01 \ mol \ P }{ 30.97 }$

$0.03261220536 \ mol \ P$

2. Convert Moles to Atoms

To convert from moles to atoms, we use Avogadro's Number.

6.022*10²³

This tells us the number of particles in 1 mole. In this case, it is atoms of phosphorus in 1 mole of phosphorus. We can create a fraction.

$\frac{6.022 *10^{23} \ atoms \ P }{1 \ mol \ P}$

Multiply by the number of moles we calculated.

$0.03261220536 \ mol \ P*\frac{6.022 *10^{23} \ atoms \ P }{1 \ mol \ P}$

The moles of phosporus will cancel.

$0.03261220536 *\frac{6.022 *10^{23} \ atoms \ P }{1 }$

$0.03261220536 *{6.022 *10^{23} \ atoms \ P }$

$1.96390701*10^{22} \ atoms \ P$

This is closest to 1.97*10²² atoms, so choice B is correct.

Chemistry

What is the mass percent of carbon in propanoic acid (c2h5cooh)? (the molar mass of c = 12.01, h = 1.0079, and o = 16.00.) 1. 43.20% 2. 43.79% 3. 48.64% 4. 51.56% 5. 52.41%

Step-by-step answer

P Answered by Specialist

Mass percent is calculated by the ratio of the mass of the substance in a sample over the total mass of the sample multiplied by 100. Assuming 1 mol of the acid, we will have 74.09 g. We do as follows:

% by mass C = 3 ( 12.01) / 74.09 x 100 = 48.63% C

Chemistry

Find the molar mass of each gas measured at the specified conditions. a) 3.48 g of gas occupying 1.52 L at 265 K and 1.63 atm. Answer in units of g/mol.

Step-by-step answer

P Answered by PhD

Molar mass = 31.64 g/mol

Explanation:

Given data:

Mass of gas = 3.48 g

Volume of gas = 1.52 L

Temperature of gas = 265 K

Pressure of gas = 1.63 atm

Molar mass of gas = ?

Solution:

Formula:

Molar mass = Mass/ number of moles

First of all we will calculate the number of moles from given data.

PV = nRT

P = pressure

V = volume

n = number of mole

R = ideal gas constant

T = Temperature

n = PV/RT

n = 1.63 atm × 1.52 L / 0.0821 atm. L /mol.K ×265 K

n = 2.48 / 21.76/mol

n = 0.11 mol

Molar mass:

Molar mass = Mass/ number of moles

Molar mass = 3.48 g / 0.11 mol

Molar mass = 31.64 g/mol

Chemistry

What is the mass percent of carbon in propanoic acid (c2h5cooh)? (the molar mass of c = 12.01, h = 1.0079, and o = 16.00.) 1. 43.20% 2. 43.79% 3. 48.64% 4. 51.56% 5. 52.41%

Step-by-step answer

P Answered by Master

Chemistry

8.50g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 176./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: Product mass Carbon dioxide 12.75g Water 3.48g Use this information to find the molecular formula of X.

Step-by-step answer

P Answered by PhD

Molecular formula = $C_{3} H_{4} O_{8}$

Explanation:

CO2 -

12.75g / 44g/mol = 0.29

H2O -

3.48g/18g/mol = 0.19

(dividing through by the smallest mole)

Ratio of CO2 to H2O = 1.5 : 1

$[(CO_{2} )_{1.5} + (H_{2} O)_{1}]x$ = 176

[12 × 1.5 + 16×2×1.5 + 1×2 + 16)x = 176

(18 + 48 + 2 + 16)x = 176

x = 176/84 = 2

⇒ Molecular formula = $C_{3} H_{4} O_{8}$

Check: Molar mass of C3H4O8 = 176g/mol.

Chemistry

Step-by-step answer

P Answered by PhD

Molecular formula = $C_{3} H_{4} O_{8}$

Explanation:

CO2 -

12.75g / 44g/mol = 0.29

H2O -

3.48g/18g/mol = 0.19

(dividing through by the smallest mole)

Ratio of CO2 to H2O = 1.5 : 1

$[(CO_{2} )_{1.5} + (H_{2} O)_{1}]x$ = 176

[12 × 1.5 + 16×2×1.5 + 1×2 + 16)x = 176

(18 + 48 + 2 + 16)x = 176

x = 176/84 = 2

⇒ Molecular formula = $C_{3} H_{4} O_{8}$

Check: Molar mass of C3H4O8 = 176g/mol.

Chemistry

Compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of CO2. Another sample weighing 5.00 g was found to contain 2.54 g of fluorine. The molar mass is found to be 448.4 g/mol. What are its empirical and molecular formulas? (AW(amu): C = 12.01, H = 1.008, F = 19.00)

Step-by-step answer

P Answered by Master

The empirical and molecular formula of the compound is CH_2F and $C_{14}H_{28}F_{14}$ respectively

Explanation:

We are given:

Mass of CO_2=3.926g

For calculating the mass of carbon:

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

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$ ......(1)

For Carbon:

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

For Fluorine:

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

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 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 1 : 2 : 1

The empirical formula for the given compound is CH_2F

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 = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

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

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is CH_2F and $C_{14}H_{28}F_{14}$ respectively

Chemistry

Step-by-step answer

P Answered by Specialist

The empirical and molecular formula of the compound is CH_2F and $C_{14}H_{28}F_{14}$ respectively

Explanation:

We are given:

Mass of CO_2=3.926g

For calculating the mass of carbon:

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

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$ ......(1)

For Carbon:

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

For Fluorine:

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

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 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 1 : 2 : 1

The empirical formula for the given compound is CH_2F

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 = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

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

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is CH_2F and $C_{14}H_{28}F_{14}$ respectively

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What is the molarity of the following solution? 3.48 mol in 4.1 L of solution

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