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3 years ago

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How many moles of calcium hydroxide are needed to produce 73.7 milliliters of water, if the density of water is 0.987 g/mL? Show

all steps of your calculation as well as the final answer. HCl + Ca(OH) 2 → H2O + CaCl2

1 answer:

kow [346]3 years ago

3 0

We first balance the equation. It will now become:
2HCl + Ca(OH)₂ ⇒ 2H₂O + CaCl₂

We also calculate the molar mass of H₂O:
2 x H = 1 g x 2 = 2g
O = 16 g
molar mass of H₂O = 18 g/mol

Then using dimensional analysis:
73.7 mL H₂O x (0.987 g H₂O/1 mL H₂O) x (1 mol H₂O/18 g H₂O)
x (1 mol Ca(OH)₂/2 mol H₂O)

= 2.02 mol Ca(OH)₂

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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

What is the concentration of FeCl3 in a solution prepared

Zinaida [17]

Well it’s the first answer couches

8 0

3 years ago

A chunk of copper has a volume of 1.45 mL and a mass of 12.93 grams. What is the density of the chunk of copper?

natta225 [31]

Answer:

$Density = 8.92 \ g/mL$

Explanation:

The equation for density is:

$Density = \frac{mass}{volume}$

We plug in the given values:

$Density = \frac{12.93 \ g}{1.45 \ mL}$

$Density = 8.92 \ g/mL$

7 0

3 years ago

Read 2 more answers

can a light object that was hit with a small force accelerate as rapidly as a heavier object hit with a big force? why or why no

Tcecarenko [31]

The acceleration is defined by force divided by the mass of the object. So, When the smaller object is hit by a small force, it can produce equal acceleration which is same as that of the bigger body hit with large force.

<h3><u>Explanation:</u></h3>

Force is defined as the product of the mass of the body its applied to and the acceleration of the body in the direction of the force. So acceleration is force divided by the mass of the body.

Let the mass of the smaller body be m and that of the larger body be M.

The smaller force applied on the smaller body be f and the larger force applied on the larger body be F.

So acceleration of the larger body = F/M.

Acceleration of the smaller body = f/m.

For the accelerations to be same,

F/M = f/m.

Or F/f = M/m.

So when the ratio of the force applied on two bodies is in ratio of their masses, the acceleration becomes equal.

3 0

3 years ago

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

<u>Step 1: </u>Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

<u>Step 2:</u> The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

<u>Step 3:</u> Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

<u>Step 4:</u> Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

<u>Step 5: </u>Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

<u>Step 6:</u> Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

<u>Step 7</u>: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

<u>Step 8:</u> Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0

3 years ago