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MatroZZZ [7]
3 years ago
4

A laboratory assistant needs to prepare 35.2 liters of hydrogen at 25.0°C and 101.3 kilopascals. This is the equation for the re

action:
2HCl + Ca → H2 + CaCl2

What volume of 2.3 M hydrochloric acid is required to produce this much gas? Use the ideal gas resource.

A. 0.625 L
B. 0.876 L
C. 1.18 L
D. 1.25 L

(Some chart that I don't understand)

Chemistry
2 answers:
Mars2501 [29]3 years ago
6 0

Answer:

Option A= 0.625 L

Explanation:

Given data:

Volume of hydrogen =35.2 L

Temperature = 25 °C = 25 + 273 = 298 K

Pressure = 101.3 Kpa = 0.99975327 atm

Molarity of HCl = 2.3 mol / L

Volume of HCl require = ?

Solution:

Formula:

PV = nRT

n = PV / RT

n = (0.99975327 atm × 35.27 L) / 0.0821 atm. L/ mol.K × 298 k

n = 35.19 atm . L / 24.4658 L. atm /mol

n = 1.438 mol

we know that molarity is equal to:

Molarity = number of moles / volume in liter

Volume in liter = Number of moles / Molarity

Volume in liter = 1.438 mol / 2.3 mol/ L

Volume in liter = 0.625 L

Ray Of Light [21]3 years ago
6 0

Answer:

D. 1.25

Explanation:

for plato users

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How would you prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0% by mass with a density of 1.19 g/mL
hoa [83]
<h2>Step 1 : Identify the given </h2>

Volume = 250mL

Density = 1.19 g/ML

<h2>Step 2 . Calculate the mass of HCL </h2>

Density = mass/volume

∴Mass = Density * Volume

= 1.19g/mL* 250mL

= 297,5g

<h2>Step 3 : Calculate the total mass of the solution, given that concentration HCL is 38% </h2>

Mass of the total solution can be calculated by the following :

38% = Mc /297.5 * 100

Mc = 38/100 *297.5

= 113.05grams

• Finally, this means that mass of the total solution of 0.125M HCL i,s 113grams, ,you would use this mass to prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0%

6 0
1 year ago
25 points please help
maxonik [38]

Answer : Option (A) Accelerator 2 model has the lowest percentage of energy lost as waste.

Solution : Given,

For Accelerator 1 model,

Input energy = 2078.3 J

Wasted energy = 663.1 J

Output energy = 1415.2 J

For Accelerator 2 model,

Input energy = 7690.0 J

Wasted energy = 2337.5 J

Output energy = 5353.5 J

For Accelerator 3 model,

Input energy = 4061.9 J

Wasted energy = 2259.6 J

Output energy = 1802.3 J

Formula used for lowest percentage of energy lost as waste is:

% energy lost as waste = (Total energy wasted / Total input energy )  ×  100

For Accelerator 1 model,

% energy lost as waste = \frac{663.1}{2078.3}\times100 = 31.90%

For Accelerator 2 model,

% energy lost as waste = \frac{2337.5}{7690.0}\times100 = 30.39%

For Accelerator 3 model,

% energy lost as waste = \frac{2259.6}{4061.9}\times100 = 55.62%

So, we conclude that the Accelerator 2 model has the lowest percentage of energy lost as waste.



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