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oksano4ka [1.4K]
3 years ago
15

What volume of 1.25M HCl would be required to prepare 180 mL of a 0.500M HCl solution?

Chemistry
2 answers:
irakobra [83]3 years ago
8 0

Answer:

Volume of 1.25 M HCl required = 72 mL.

Explanation:

This is a dilution concept.

It is that if we dilute a solution, the number of moles of solute will remain the same in the diluted solution. It is the moles per unit volume which changes on dilution.

So after dilution the moles of HCl present will be :

Moles = molarity X volume = 0.500 X 180 mL = 0.500 X 0.18 L = 0.09 moles

this is the moles present before dilution

moles before dilution required = 0.09 = molarity X volume

volume of concentrated solution required = moles / molarity = 0.09 / 1.25

volume of 1.25 M solution required = 0.072 L = 72mL

Vlad1618 [11]3 years ago
7 0
The volume required of 1.25M HCL to prepare 180ml of a 0.500M Hcl solution is 72ml

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What volume of carbon dioxide (CO2) will be produced if 2.90 moles of it on (Fe) is produced?
dolphi86 [110]

Answer:

= 97.44 Liters at S.T.P

Explanation:

The reaction between Iron (iii) oxide and Carbon monoxide is given by the equation;

Fe2O3(s)+ 3CO(g) → 3CO2(g) + 2Fe(s)

From the reaction when the reactants react, 2 moles of Fe and 3 moles of CO2 are produced.

Therefore; Mole ratio of Iron : Carbon dioxide is 2:3

Thus; Moles of Carbon dioxide = (2.9/2)×3

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Therefore;

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6 0
3 years ago
Which gas law states that the pressure of a gas decreases when volume is increased and the temperature is unchanged
Nesterboy [21]

Answer: Boyle's law

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P∝\frac{1}{V}       at constant temperature

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3 0
4 years ago
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kozerog [31]

Answer:

A

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Consider the reaction: 2 H2O (g)-->2 H2 (g) + O2 (g). ΔH=483.6 Kj/mol. If 2 moles of H2O (g) are converted H2(g) and O2(g) ag
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DE = dH - PdV 

<span>2 H2O(g) → 2 H2(g) + O2(g) </span>

<span>You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. </span>

<span>1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. </span>

<span>Since it is done under constant temp and pressure that means the volume change will be equal to the volume of 1 mole of gas </span>

<span>2 moles reacts to form 3 moles </span>

<span>The gas equation is </span>

<span>PV = nRT </span>
<span>P = pressure </span>
<span>V = volume (unknown) </span>
<span>n = moles (1) </span>
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<span>- the gas constant is different for different units of temp and pressure (see wikki link) in this case temp and pressure are constant, and we want to put the result in an equation that has Joules in it, so we select 8.314 JK^-1mol^-1) </span>
<span>T = temp in Kelvin (kelvin = deg C + 273.15 </span>
<span>So T = 403.15 K </span>

<span>Now, you can see that PV is on one side of the equation, and we are looking to put PdV in our dE equation. So we can say </span>

<span>dE = dH -dnRT (because PV = nRT) </span>

<span>Also, since the gas constant is in the unit of Joules, we need to convert dH to Joules </span>

<span>dH = 483.6 kJ/mol = 483600 Joules/mol </span>

<span>dE = 483600 J/mol - (1.0 mol x 8.314 J mol^-1K-1 x 403.15 K) </span>
<span>dE = 483600 J/mol - 3351.77 J </span>
<span>dE = 480248.23 J/mol </span>
<span>dE = 480.2 kJ/mol </span>
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Answer:

mnbnm

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