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LuckyWell [14K]
2 years ago
7

The psychoactive drug sold as a methamphetamine - "speed" (c10h15n) - undergoes a series of reactions in the body; the net resul

t of these reactions is the oxidation of solid methamphetamine by oxygen gas to produce carbon dioxide gas, liquid water, and nitrogen gas. what is the coefficient of nitrogen in the balanced equation for the reaction? (balance the equation with the smallest possible whole number coefficients.)
Chemistry
1 answer:
Phoenix [80]2 years ago
8 0

Answer: The coefficient of nitrogen in the given equation is 2.

Explanation: The reaction for the oxidation of methamphentamine with oxygen gas in the body is given by:

4C_{10}H_{15}N(s)+55O_2(g)\rightarrow 40CO_2(g)+30H_2O(l)+2N_2(g)

By Stoichiometry,

4 moles of methamphentamine reacts with 55 moles of oxygen gas to produce 40 moles of carbon dioxide gas, 30 moles of water and 2 moles of nitrogen gas.

Coefficient of C_{10}H_{15}N=4

Coefficient of O_2=55

Coefficient of CO_2=40

Coefficient of H_2O=30

Coefficient of N_2=2

Hence, the coefficient of nitrogen in the given equation is 2.

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The balanced chemical reaction is written as:

MnCl2(s)+H2SO4(aq)→MnSO4(aq)+2HCl(g)

We are given the amount of hydrochloric acid to be produced in the reaction. This value will be the starting point for the calculations.We do as follows:

.0525 L HCl ( 1 mol / 22.4 L ) ( 1 mol MnCl2 / 2 mol HCl ) (125.84 g / 1 mol ) = 0.15 g MnCl2
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3 years ago
Solid mixture contains MgCl2 and NaCl. When 0.5000 g of this solid is dissolved in enough water to form 1.000 L of solution, the
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Answer:

83,40 (w/w) %

Explanation:

The osmotic pressure (π) is defined as:

π = iMRT

Where i is Van't Hoff factor (3 for MgCl₂ and 2 for NaCl), M is molarity, R is gas contant (0,082atmL/molK) and T is temperature (298,15K)

iM = π / RT

iM = 0,01607mol/L

It is possible to write:

<em>3x+2y = 0,01607mol/L </em><em>(1)</em>

Where x are moles of MgCl₂ and y moles of NaCl.

<em>-M = moles of each compund because M is molarity (moles/L) and there is 1,000L-</em>

Knowing molar mass of MgCl₂ is 95,2 g/mol and for NaCl is 58,44g/mol:

<em>x×95,211 + y×58,44 = 0,5000g </em><em>(2)</em>

Replacing (2) in (1):

3x+2(0,0086 - 1,629x) = 0,01607mol/L

-0,258x = -0,00113

<em>x = 0,004380 moles of MgCl₂</em>

In grams:

0,004380 moles of MgCl₂×(95,211g/mol) = 0,417g of MgCl₂

Mass percent is:

(0,4170g of MgCl₂/0,5000g of solid) ×100 = <em>83,40 (w/w) %</em>

I hope it helps!

4 0
3 years ago
How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo
riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)

The expression used will be:

\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

where,

\Delta H = heat released by the reaction = ?

m = mass of benzene = 125 g

c_{p,g} = specific heat of gaseous benzene = 1.06J/g^oC

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC

\Delta H_{vap} = enthalpy change for vaporization = 33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]

\Delta H=-67682.5J=-67.7kJ

Therefore, the energy removed must be, -67.7 kJ

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