Chemistry, 02.10.2019 22:00 destinytofell4630

One liter of 0.1 m tris buffer is prepared and adjusted to ph of 8.2.
(a) write the equation of the acid dissociation of tris-hcl identifying the weak acid and the conjugate base in your equation. (the structure of tris-hcl salt is shown on the lecture slides and pka=8.30).

(b) what is the effective buffering range of a tris buffer?

(c) what are the concentrations of the conjugate acid and weak base at ph=8.2?

(d) what is the ratio of conjugate base to weak acid if 1.5 ml of 3.0 m hcl is added to 1.0 l of the buffer? what is the new ph? has the buffer functioned effectively? compare the ph change to that of problem 9a in which the same amount of acid was added to the same volume of pure water.

(e) what is the ratio of conjugate base to weak acid if 1.5 ml of 3.0 m naoh is added to 1.0 l of the buffer? what is the new ph? has the buffer functioned effectively? compare the ph change to that of problem 3b in which the same amount of base was added to the same volume of pure water.

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Answer from: Quest

the best answer is d because the strong cohesion is caused by the ability of water molecules to undergo hydrogen bonding.

a is incorrect because when water solidifies, it becomes ice, which expands and occupies a bigger volume for the same mass of water in its liquid state.

b is incorrect because water has high surface tension. besides bonding to itself it hydrogen bonds to other molecules at the top decreasing possibility for breaking of attractive forces.

c is incorrect because the water molecule has oxygen which is highly electronegative and so pulls the electron pair towards itself creating a dipole moment, making the molecule polar.

e is incorrect because water has strong adhesion, ability to stick to other molecules such as glass because of this molecules having stronger hydrogen bonding than water.

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³

Answer from: Quest

At a given pressure, gas particles with smaller masses move faster.

Answer from: Quest

option c: 1605.1 kj

the balanced chemical reaction for combustion of methane is as follows:

$ch_{4}+2o_{2}\rightarrow co_{2}+2h_{2}o$

here, enthalpy formation of $ch_{4}, o_{2}, co_{2} and h_{2}o$ is -74.7 kj/mol, 0 kj/mol, -393.5 kj/mol and -241.82 kj/mol respectively.

the heat released or $\delta h_{reaction}$ can be calculated as follow:

$\delta h_{reaction}=\delta h_{f}_{co_{2}}+2(\delta h_{f}_{h_{2}o} h_{f}_{ch_{4}}-2\delta h_{f}_{o_{2}})$

putting the values,

$\delta h_{reaction}=-393.5 kj/mol+(.82 kj/.6-(2)(0 kj/mol))=802.54 kj/mol$

thus, heat released by combustion of 1 mol of methane is 802.54 kj/mol. thus, for 2 mol, heat released will be:

$\delta h_{reaction}=(2 mol)(802.54 kj/mol)=1605.1 kj$

therefore, heat released by the combustion of 2 mol of methane is 1605.1 kj.

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One liter of 0.1 m tris buffer is prepared and adjusted to ph of 8.2.
(a) write the equation o...

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