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Alenkinab [10]
2 years ago
8

A stock solution of ammonia in water is 28 wt% NH3. The Formal Weight (FW) of NH3 is 17.03 g/mol). The solution density at room

temperature is 0.90 g/mL. What is the molarity of the solution in units of mol/L
Chemistry
1 answer:
weqwewe [10]2 years ago
4 0

Answer:

[NH₃] = 14.7 mol/L

Explanation:

28 wt% is a type of concentration that indicates that 28 g of ammonia is contained in 100 g of solution.

Let's determine the amount of ammonia:

28 g . 1 mol / 17.03g = 1.64 moles of NH₃

You need to consider that, when you have density's data it is always referred to solution:

Mass of solution is 100 g, let's find out the volume

0.90 g/mL = 100 g /V

V = 100 g / 0.90mL/g → 111.1 mL

We convert the volume to L → 111.1 mL . 1 L/1000mL = 0.1111 L

mol/L = 1.64 mol/0.1111L → 14.7 M

mol/L = M → molarity a sort of concentration that indicates the moles of solute in 1L of solution

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One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature af
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Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

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3 years ago
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Answer is in the file below

tinyurl.com/wpazsebu

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2 years ago
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Answer:

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