Why do (or can) your muscles hurt after a good workout

Consider the following reaction. SO2Cl2 → SO2 + Cl2. After collecting experimental data you found that plotting ln[SO2Cl2] vs. t

Nikitich [7]

Answer:

$[SO_2Cl_2]_{600}= 0.0842 M$

Explanation:

Some theoretical knowledge is required here. We should understand that whenever we plot the natural logarithm, ln, of a concentration vs. time and obtain a straight line, this indicates a first-order reaction. That said, since this is the case here, we have a first-order reaction with respect to $SO_2Cl_2$ .

The linear equation has the following terms:

$y = -0.000290t - 2.30$

It is a linear form of the integrated first-order law equation:

$ln[SO_2Cl_2]_t = -kt + ln[SO_2Cl_2]_o$

Therefore, the rate constant, k, is:

$k = 0.000290 s^{-1}$

The natural logarithm of initial molarity is:

$ln[SO_2Cl_2]_o = -2.30$

Using the equation, we may substitute for t = 600 s and obtain the natural logarithm of the concentration at that time:

$ln[SO_2Cl_2]_{600} = -0.000290 s^{-1}\cdot 600 s - 2.30 = -2.474$

Take the antilog of both sides to find the actual molarity:

$[SO_2Cl_2]_{600}=e^{-2.474} = 0.0842 M$

4 0

2 years ago

List 4 things that could happen to the water used in watering local golf course

Advocard [28]

The water used to water a golf course may be:
1) Carried off the surface of the golf course into nearby streams or rivers
2) Evaporated
3) Lost to the ground by the process of leaching
4) Absorbed by the vegetation in the golf course such as the grass, trees and shrubs.

3 0

3 years ago

One method used commercially to peel potatoes is to soak them in a solution of NaOH for a short time and then remove them from t

sergij07 [2.7K]

Answer:

b. 2.28 M

Explanation:

The reaction of neutralization of NaOH with H2SO4 is:

2NaOH + H2SO4 → Na2SO4 + 2H2O

<em>Where 2 moles of NaOH react per mole of H2SO4</em>

<em />

To solve the concentration of NaOH we need to find the moles of H2SO4. Using the chemical equation we can find the moles of NaOH that react and with the volume the molar concentration as follows:

<em>Moles H2SO4:</em>

45.7mL = 0.0457L * (0.500mol/L) = 0.02285 moles H2SO4

<em>Moles NaOH:</em>

0.02285 moles H2SO4 * (2moles NaOH / 1 mol H2SO4) = 0.0457moles NaOH

<em>Molarity NaOH:</em>

0.0457moles NaOH / 0.020L =

2.28M

Right option:

7 0

2 years ago

Cinnamon owes its flavor and odor to cinnamaldehyde (C9H8O). Determine the boiling point elevation of a solution of 92.7 mg of c

Flauer [41]

Answer:

The boiling point elevation is 3.53 °C

Explanation:

∆Tb = Kb × m

∆Tb is the boiling point elevation of the solution

Kb is the molal boiling point elevation constant of CCl4 = 5.03 °C/m

m is the molality of the solution is given by moles of solute (C9H8O) divided by mass of solvent (CCl4) in kilogram

Moles of solute = mass/MW =

mass = 92.7 mg = 92.7/1000 = 0.0927 g

MW = 132 g/mol

Moles of solute = 0.0927/132 = 7.02×10^-4 mol

Mass of solvent = 1 g = 1/1000 = 0.001 kg

m = 7.02×10^-4 mol ÷ 0.001 kg = 0.702 mol/kg

∆Tb = 5.03 × 0.702 = 3.53 °C (to 2 decimal places)

6 0

2 years ago

A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the

Tresset [83]

Answer:

a. Molarity= $M =2.1x10^{-1}M$

b. Molality= $m=2.0x10^{-1}m$

Explanation:

Hello,

In this case, given the information about the aniline, whose molar mass is 93g/mol, one could assume the volume of the solution is just 200 mL (0.200 L) as no volume change is observed when mixing, therefore, the molarity results:

$M=\frac{n_{solute}}{V_{solution}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L} =2.1x10^{-1}M$

Moreover, the molality:

$m=\frac{n_{solute}}{m_{solvent}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L*\frac{1.05kg}{1L} } =2.0x10^{-1}m$

Best regards.

7 0

2 years ago