Consider the following reaction:2 HgCl2(aq) + C2O42-(aq) 2 Cl-(aq) + 2 CO2(g) + Hg2Cl2(s)(a) The rate law for this reaction is first order in HgCl2(aq) and second order in C2O42-(aq). What is the rate law for this reaction?Rate = k [HgCl2(aq)] [C2O42-(aq)]Rate = k [HgCl2(aq)]2 [C2O42-(aq)]Rate = k [HgCl2(aq)] [C2O42-(aq)]2Rate = k [HgCl2(aq)]2 [C2O42-(aq)]2Rate = k [HgCl2(aq)] [C2O42-(aq)]3Rate = k [HgCl2(aq)]4 [C2O42-(aq)](b) If the rate constant for this reaction at a certain temperature is 0.0169, what is the reaction rate when [HgCl2(aq)] = 0.158 M and [C2O42-(aq)] = 0.202 M?Rate = M/s.(c) What is the reaction rate when the concentration of HgCl2(aq) is doubled, to 0.316 M while the concentration of C2O42-(aq) is 0.202 M?Rate = M/s

Answer:

The Universe is constantly expanding and as it expands the stars and objects in space move farther apart, just like the points on the balloon when air is blown into it. Density and distribution of "stars" as the balloon expands because when volume increases the density will increase.

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Answer:

Evaporation is a technique used to separate out homogeneous mixtures where there is one or more dissolved salts.

Explanation:

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Answer:

m = 0.00659 kg = 6.59 g

Explanation:

From Faraday's Law of Electrolysis, we know that:

m = ZQ

where,

m = mass of silver deposited = ?

Q = charge supplied = (1.65 A-hr)(3600 s/1 hr) = 5940 C

Z = electrochemical equivalent of silver = 1.18 x 10⁻⁶ kg/C

Therefore,

m = (1.11 x 10⁻⁶ kg/C)(5940 C)

m = 0.00659 kg = 6.59 g

The mass of silver that would be deposited by a battery is 6.65 grams

The precipitation of Ag requires the removal of one electron. The reduction process for silver electrode at the cathode is as follows:

[tex]\mathbf{Ag^+ + e^- \to Ag(s)}[/tex]

∴

In one mole of an electron, the charge carried = 96500 C

Recall that:

The atomic mass of silver (Ag) = 108 g

∴

The mass of silver that would be deposited in a 5940 C can be computed as:

[tex]\mathbf{=5940\ C \times \dfrac{108 \ g }{96500 \ C}}[/tex]

= 6.65 grams

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Since the forces equal to 0, it's balanced because the object isn't moving.

Answer:

it has 12 carbon atoms and 22 hydrogen atoms and 11 oxygen atoms

Explanation:

Answer:

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Explanation:

Answer:

309 K

Explanation:

Step 1: Convert the pressure to atm

We will use the conversion factor 1 atm = 760 mmHg.

732 mmHg × 1 atm/760 mmHg = 0.963 atm

Step 2: Calculate the moles (n) of the ideal gas

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.963 atm × 8.4 L/0.0821 atm.L/mol.K × 298 K

n = 0.33 mol

Step 3: Calculate the temperature change

We will use the following expression.

Q = n × Cp × ΔT

ΔT = Q/n × Cp

ΔT = 75 J/0.33 mol × 20.79 J/mol.K

ΔT = 11 K

Step 4: Calculate the final temperature

T = 298 K + 11 K = 309 K

Answer:  The volume occupied at an altitude of 20.0 km is 34.5289 L

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]     (At constant temperature and number of moles)

where,

[tex]P_1[/tex] = initial pressure of gas = 101.325 kPa ( sea level)

[tex]P_2[/tex] = final pressure of gas = 10 kPa

[tex]V_1[/tex] = initial volume of gas = 3.40774 L

[tex]V_2[/tex] = final volume of gas = ?

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]101.325\times 3.40774=10\times V_2[/tex]

[tex]V_2=34.5289L[/tex]

Therefore, the volume occupied at an altitude of 20.0 km is 34.5289 L

Answer:

a. PbSO4, Ksp = 1.7x10^-8.

Explanation:

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In this case, since the solubility product indicates how likely a solid is able to ionize and consequently dissolve in water, we can infer that the larger the solubility product Ksp, the more ions are able dissolve in water; therefore the proper answer goes with the largest Ksp, which is a. PbSO4, Ksp = 1.7x10^-8 since the power goes closer to 1 than the other options.

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Answer:

The temperature is on the y-axis and the time is on the x-axis.

Explanation:

Which statement describes how phase changes can be diagrammed as a substance is heated? (D is the answer)

The phase is on the y-axis and the temperature is on the x-axis.

The temperature is on the y-axis and the phase is on the x-axis.

The time is on the y-axis and the temperature is on the x-axis.

The temperature is on the y-axis and the time is on the x-axis.

Which statement describes the appearance of a temperature-vs.-time graph? (C is the answer)

A horizontal line shows that the temperature increases at a constant rate over time.

A vertical line shows that the temperature decreases at a constant rate over time.

Horizontal lines where the temperature is constant during phase changes connect upward-sloping lines where the temperature increases.

Horizontal lines where the temperature increases are connected by upward-sloping lines where the temperature is constant for each phase.