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3 years ago

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A solution of rubbing alcohol is 68.6 % (v/v) isopropanol in water. How many milliliters of isopropanol are in a 88.2 mL sample

of the rubbing alcohol solution? Express your answer to three significant figures.

1 answer:

KonstantinChe [14]3 years ago

8 0

From the information given, the total volume of rubbing alcohol is 88.2 ml

68.6 % of this volume is isopropanol.

We will assume 88.2 ml represents 100% volume, so the volume of water will be 31.4 %

The volume of isopropanol is

68.6/100 x 88.2 → 0.686 × 88.2 = 60.505 ml

The volume of isopropanol is 60.5 ml.

Volume of water will be 88.20 - 60.5 = 27.7 ml

(27.7 / 88.2 × 100 = 31.4% )

Adding 60.5 ml of isopropanol to 27.7 ml of water to make up 88.2 ml will give 68.6 % v/v isopropanol to water solution.

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Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

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Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

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Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

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Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

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W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

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3 years ago

For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e

sergiy2304 [10]

Answer:a) 11.34 g of ethane $(C_2H_6)$ can be formed

b) $C_2H_4$ is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$

1. $\text{Moles of} H_2=\frac{4.21}{2}=2.10moles$

2. $\text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ require 1 mole of $H_2$

Thus 0.378 moles of $C_2H_4$ will require= $\frac{1}{1}\times 0.378=0.378moles$ of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent.

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As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.378 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.378=0.378moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g$

Thus 11.34 g of ethane is formed.

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QveST [7]

Answer:

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Here was the process:

CaBr2+2Li3PO4 -> Ca3(PO4)2+LiBr

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3CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr

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3 years ago

Which chemical equation shows that the total mass during a chemical reaction stays the same?

k0ka [10]

Answer:

Option A) <u><em>Mg + Cl₂ → MgCl₂</em></u>

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The law of conservation of mass is guaranteed in a chemical equation. Since the mass of the atoms do not change, that means that the number of each kind of atoms in the reactant side is equal to the number of atoms of the same kind in the product side.

The first equation is:

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<u />

Number of atoms:

atom Reactant side Product side

Mg 1 1

Cl 2 2

Therefore, the table displays that there are the same number of atoms of each kind on both sides, showing that<em> the total mass during the chemical reaction stays the same.</em>

<u />

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<em />

<u><em>C) 2Na + 2H₂O → NaOH + H₂</em></u>

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5 0

2 years ago

Read 2 more answers