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algol13
2 months ago
9

A 1.34 mole sample of LiCl dissolves in water, The volume of the final

Chemistry
1 answer:
eduard [2.7K]2 months ago
0 0

Response:

1.56 \,\,mol/L

Explanation:

Molarity is defined as the number of moles of a solute contained in one liter of solution.

Molarity = volume of the solution in liters/number of moles of solute present in the solution

The volume of the solution is 0.86 L

Additionally, a sample of 1.34 moles of LiCl dissolves in the solvent

Thus,

Molarity of the Solution =

\frac{1.34}{0.86}=1.56 \,\,mol/L

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1.562 g sample of the alcohol CH3CHOHCH2CH3 is burned in an excess of oxygen. What masses of H2O and CO2 should be obtained
eduard [2782]

Answer:

m_{CO_2}=3.709gCO_2 \\\\m_{H_2O}=1.898gH_2O

Explanation:

Hello.

In this scenario, as the molecular formula for the specified alcohol is C₄H₁₀O (with a molar mass of 74.14 g/mol), its combustion reaction can be represented as follows:

C_4H_1_0O+6O_2\rightarrow 4CO_2+5H_2O

This implies a mole ratio of 1:4 with carbon dioxide (molar mass = 44.01 g/mol) and a mole ratio of 1:5 with water (molar mass = 18.02 g/mol), allowing us to determine the resultant masses as follows:

m_{CO_2}=1.562gC_4H_1_0O*\frac{1mol}{74.14gC_4H_1_0O} *\frac{4molCO_2}{1molC_4H_1_0O} *\frac{44.01gCO_2}{1molCO_2}=3.709gCO_2 \\\\m_{H_2O}=1.562gC_4H_1_0O*\frac{1mol}{74.14gC_4H_1_0O} *\frac{5molH_2O}{1molC_4H_1_0O} *\frac{18.02gH_2O}{1molH_2O}=1.898gH_2O

Best regards!

7 0
2 months ago
Two glasses labeled A and B contain equal amounts of water at different temperatures. Kim put an antacid tablet into each of the
eduard [2782]

Answer:

The correct statement is D. The water in Glass A is at a lower temperature compared to Glass B; consequently, the particles in Glass A exhibit reduced movement.

Explanation:

Raising the temperature increases the solubility of solutes.

The experiment indicates that's glass B is at a higher temperature than glass A since the antacid dissolves more quickly in glass B than in glass A. Therefore, glass A must be cooler, leading to slower particle movement compared to glass B.

6 0
3 months ago
Read 2 more answers
One type of breathalyzer employs a fuel cell to measure the quantity of alcohol in the breath. When a suspect blows into the bre
Alekssandra [3086]

Response:The ethanol percentage is 0.1093%

Explanation:

As given:

t = time = 10 s

I = current = 320 mA

F = Faraday's constant = 96485.3365 C mol⁻¹

n = number of electrons transferred = 4

Molecular weight of ethanol is 46 g/mol

Question: What is the percent (by volume) of ethanol in the driver's breath, %E =?

First, calculate the ethanol mass:

W=\frac{\frac{46}{4} *0.32*10}{96485.3365} =3.814x10^{-4} g

The moles of ethanol:

n_{ethanol} =3.814x10^{-4} g*\frac{1mol}{46} =8.291x10^{-6} moles

Applying the ideal gas law formula:

V=\frac{nRT}{P}

Here:

T = 26°C = 299 K

P = 1 atm

Substituting in the values:

V=\frac{8.291x10^{-6}*0.082*299 }{1} =2.033x10^{-4} L=0.2033mL

The percentage of ethanol:

E=\frac{0.2033}{186} *100=0.1093%

3 0
2 months ago
How much volume (in cm3) is gained by a person who gains 11.1 lbs of pure fat?
Anarel [2989]

The density of human fat is 0.918 g/cm^{3}. The mass of the pure fat is 11.1 lbs.

First, convert the mass from pounds to grams as follows:

1 lb=453.592 g

Thus,

11.1 lb\rightarrow 11.1\times 453.592 g=5034.875 g

Density is defined as mass per unit volume, meaning volume can be calculated as:

V=\frac{m}{d}

By substituting the values,

V=\frac{5034.875 g}{0.918 g/cm^{3}}=5484.61 cm^{3}

Consequently, the volume gained by the individual will be 5484.61 cm^{3}.

6 0
2 months ago
Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr
eduard [2782]

Answer:

The integer value of x in the hydrate is 10.

Explanation:

Molarity=\frac{Moles}{Volume(L)}

Molar concentration of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

0.0366 M=\frac{n}{5.00 L}

n=0.0366 M\times 5 mol=0.183 mol

Weight of hydrated sodium carbonate = n = 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol + x * 18 g/mol

n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}

0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}

106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}

By solving for x, we arrive at:

x = 9.95, approximating to 10

The integer x in the hydrate equals 10.

6 0
2 months ago
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