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alekssr [168]
2 years ago
9

The graph above shows the changes in temperature recorded for the 2.00 l of h2o surrounding a constant-volume container in which

a 1.00 g sample of benzoic acid was combusted. Assume that heat was not absorbed by the container or lost to the surroundings

Chemistry
1 answer:
Otrada [13]2 years ago
6 0

Answer:

25.2 kJ

Explanation:

The complete question is presented in the attached image to this answer.

Note that, the heat gained by the 2.00 L of water to raise its temperature from the initial value to its final value comes entirely from the combustion of the benzoic acid since there are no heat losses to the containing vessel or to the environment.

So, to obtained the heat released from the combustion of benzoic acid, we just calculate the heat required to raise the temperature of the water.

Q = mCΔT

To calculate the mass of water,

Density = (mass)/(volume)

Mass = Density × volume

Density = 1 g/mL

Volume = 2.00 L = 2000 mL

Mass = 1 × 2000 = 2000 g

C = specific heat capacity of water = 4.2 J/g.°C

ΔT = (final temperature) - (Initial temperature)

From the graph,

Final temperature of water = 25°C

Initial temperature of water = 22°C

ΔT = 25 - 22 = 3°C

Q = (2000×4.2×3) = 25,200 J = 25.2 kJ

Hope this Helps!!!

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So converting it into grams we have to multiply;

1.0073 amu X  1.66054 X 10^{-24} g/amu = 1.673 X 10^{-24} g

Now, Volume = 1/6πd³ as diameter is given as 1.0 X 10^{-15} m converting it to cm will require to multiply with 100

∴ Volume  = 1/6π (1.0 X 10^{-15}mX 100 cm / 1 m)^{3}

Hence, volume =  5.236 X 10^{-40} cm^{3}

Therefore, Density = mass / volume

∴ Density =  1.673 X 10^{-24} g / 5.236 X 10^{-40} cm^{3}

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Therefore two electrons must be added to the right as shown below:

MnCO3 -------> MnO2+ HCO3-+ 2e-Now,there is one negative charge HCO3- and 1 negative charge on the two electrons making a total of -3 charges on the right. There is zero charge on the left.

To balance the equation,add3H+on the right,to cancel out the charges.

MnCO3 --------> MnO2+HCO3-+2e-+3H+

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MnCO3+2H2O ------->MnO2+HCO3-+2e-+3H+

3 0
2 years ago
If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th
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Answer:

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be  the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

 [Pb²⁺] =   s

Then [I⁻] = 2s

K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} =  4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of  }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

B)

The Concentration of Pb²⁺  in water is calculated as :

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}

\mathbf{s} =\sqrt[3]{1.75*10^{-9}}

\mathbf{s} =\mathbf{1.21*10^{-3}  \ mol/L }

The Concentration of Pb²⁺  in 1.0 mol·L⁻¹ NaI

\mathbf{PbCl{_2}}  \leftrightharpoons    \ \ \ \ \ \ \  \mathbf{Pb^{2+}}   \ \ \ \  \ +   \ \  \ \ \ \ \ \mathbf{2 I^-}

                             \ \ \ \ \ \ \  \ \   \ \  \ \ \ \ \ \ \  \mathbf0}   \ \ \ \  \ \ \ \ \ \   \ \ \ \ \ \mathbf{1.0}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{+2x}

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The equilibrium constant:

K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \  m/L

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the  common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0
2 years ago
A sample of gasoline contains various hydrocarbons, which comprise atoms of carbon, hydrogen, and oxygen. The hydrocarbons mix t
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Answer:

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Explanation:

Mixture can be easily separated by physical methods. Homogeneity and heterogeneity of a mixture is determined by whether the components there in are in a single phase and evenly distributed or not.

A solution has a solute evenly dissolved in  solvent to form a liquid substance.

An element is the basic form of  substance which cannot be broke down into any other simpler unit.  

I hope this was helpful.

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