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3 years ago

How many liters of hydrogen gas is produced from 3.712 g of magnesium with 104.2ml of 1.385 mol/L HCL (aq) at SATP?

Chemistry

1 answer:

Savatey [412]3 years ago

8 0

Answer:

The correct answer is 1.61 L.

Explanation:

Based on the given information, 3.712 grams of Mg reacts with 104.2 ml of 1.385 mol per L HCl at SATP, there is a need to find the amount of hydrogen gas produced in liters.

The chemical reaction taking place in the given case is,

Mg + 2HCl = MgCl2 + H2

The reacting moles of each reactants is,

Moles of Mg = 3.712 g/24.305 g = 0.153 moles

Moles of HCl = 1.385 mol/L * 0.1042 L = 0.144 moles

From the reaction it is clear that Mg and HCl are present in 1:2 molar ratio. Therefore, 0.153 moles of Mg can completely react with 0.306 moles of HCl. However, the moles of HCl obtained in the given case is only 0.144 moles, thus, HCl is a limiting reactant.

Now the moles of hydrogen produced is,

n = 0.144 moles of HCl * (1 mole H2/2 mol HCl) = 0.072 moles

Finally to find the liters of hydrogen gas produced, the ideal gas equation is used, that is, PV = nRT

At STAP, the value of T is 273 K and pressure is 1 atm, the value of R is 0.082 atm.L/mol.L. Now putting the values we get,

PV = nRT

V = nRT/P

V = 0.072 mol * 0.082 atm.L/mol.L*273/ 1 atm

V = 1.61 L

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Since 1 mol H2O has 2 mol H, this is equivalent to (0.06043)*2 = 0.1209 mol H.

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3 years ago

Sewage and industrial pollutants dumped into a body of water can reduce the dissolved oxygen concentration and adversely affect

MArishka [77]

Answer:

FALSE

Since 0.385 < 0.526, the value for week 3 is accepted.

Explanation:

Qexp = (|Xq - Xₙ₋₁|)/w

where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data

First, the data are arranged in decreasing order, from highest to lowest:

3. 5.6

2. 5.1

8. 5.1

1. 4.9

6. 4.9

5. 4.7

7. 4.5

4. 4.3

Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3

Qexp = (|5.6 - 5.1|)/1.3 = 0.385

From tables, at 95% confidence level, for n = 8, Qcrit = 0.526

Since 0.385 < 0.526, the value for week 3 is accepted.

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3 years ago

What is the interval used on the y-axis of this graph?

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Answer:

Explanation:

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3 years ago

The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal

spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

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ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

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4 years ago

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Vikentia [17]

Answer:

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