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3 years ago

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what is the concentration (in %v/v)) of the following solutions? a. 25 mL of ethanol is diluted to a volume of 150 mL with water

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1 answer:

fenix001 [56]3 years ago

8 0

The symbol %v/v means percent by volume. Assuming there is no volume effects when these substances are mixed, we calculate as follows:

% v/v = (25 mL ethanol / 25 mL + 150 mL ) x 100
%v/v = 14.29 mL ethanol / mL solution

Hope this answers the question.

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The density of magnesium is 1.7 g/cm3, and the density of iron is 7.9 g/cm3. Consider a block of iron with a mass of 819 g. What

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Answer:

mass of Mg=0.016g

Explanation:

P(Mg)=1.7g/cm³

P(Fe)=7.9g/cm³

m(Fe)=819g

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7.9=819/v(Fe)

v(Fe)=0.0096cm³

since v(Mg)=v(Fe)

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P(Mg)=m(Mg) /v(Mg)

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What type of forces allow for an object to stay at rest or continue at constant velocity.

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Answer:Inertia is the property of a body to remain at rest or to remain in motion with constant velocity. Some objects have more inertia than others because the inertia of an object is equivalent to its mass.

Explanation:

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3 years ago

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Ammonia (NH3) clouds are present around some planets. Calculate the number of grams of ammonia produced by the reaction of 5.4 g

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Answer:

30.4 g. NH3

Explanation:

This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.

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5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)

Now, we can use the relationship between H2 and NH3.

2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3

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1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)

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2 years ago

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A reaction that had two compounds as reactants and two compounds as products is most likely a

sergejj [24]

double-replacement reaction

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3 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

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3 years ago