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Alenkasestr [34]

4 years ago

8

Which statement is true? A. A sample statistic can never be equal to the population parameter. B. Point estimates are defined fo

r both samples and populations. C. To calculate a sample proportion, we need to know the population size. D. To calculate a point estimate, we need to know the sample size

2 answers:

Allushta [10]4 years ago

5 0

i think its either a or d not sure, hope this helps

Helga [31]4 years ago

5 0

Answer:

A

Explanation:

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Consider the nitration by electrophilic aromatic substitution of salicylamide to iodosalicylamide. Reaction scheme illustrating

kvv77 [185]

<u>Answer:</u> The percent yield of the reaction is 68.68%.

<u>Explanation:</u>

To calculate the mass of salicylamide, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of salicylamide = 1.06 g/mL

Volume of salicylamide = 3.65 mL

Putting values in above equation, we get:

$1.06g/mL=\frac{\text{Mass of salicylamide}}{3.65mL}\\\\\text{Mass of salicylamide}=(1.06g/mL\times 3.65mL)=3.869g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of salicylamide = 3.869 g

Molar mass of salicylamide = 137.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of salicylamide}=\frac{3.869g}{137.14g/mol}=0.0295mol$

The chemical equation for the conversion of salicylamide to iodo-salicylamide follows:

$\text{salicylamide }+NaI+NaOCl+EtOH\rightarrow \text{iodo-salicylamide }$

By Stoichiometry of the reaction:

1 mole of salicylamide produces 1 mole of iodo-salicylamide

So, 0.0295 moles of salicylamide will produce = $\frac{1}{1}\times 0.0295=0.0295moles$ of iodo-salicylamide

Now, calculating the mass of iodo-salicylamide from equation 1, we get:

Molar mass of iodo-salicylamide = 263 g/mol

Moles of iodo-salicylamide = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of iodo-salicylamide}}{263g/mol}\\\\\text{Mass of iodo-salicylamide}=(0.0295mol\times 263g/mol)=7.76g$

To calculate the percentage yield of iodo-salicylamide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iodo-salicylamide = 5.33 g

Theoretical yield of iodo-salicylamide = 7.76 g

Putting values in above equation, we get:

$\%\text{ yield of iodo-salicylamide}=\frac{5.33g}{7.76g}\times 100\\\\\% \text{yield of iodo-salicylamide}=68.68\%$

Hence, the percent yield of the reaction is 68.68%.

4 0

4 years ago

Ephedrine, a central nervous system stimulant, is used in nasalsprays as a decongestant. this compound is a weak organic base: {

sashaice [31]

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base:

C10H15ON (aq) + H2O (l) -> C10H15ONH+ (aq) + OH- (aq)

A 0.035 M solution of ephedrine has a pH of 11.33.

a) What are the equilibrium concentrations of C10H15ON, C10H15ONH<span>+, and OH-</span>?

b) Calculate <span>Kb</span> for ephedrine.

c(C₁₀H₁₅NO) = 0,035 M.<span>
pH = 11,33.
pOH = 14 - 11,33 = 2,67.
[OH</span>⁻] = 10∧(-2,67) = 0,00213 M.<span>
[OH</span>⁻] = [C₁₀H₁₅NOH⁺] = 0,00213 M.<span>
[</span>C₁₀H₁₅NO] = 0,035 M - 0,00213 M = 0,03287 M.<span>
Kb = [OH</span>⁻] · [C₁₀H₁₅NOH⁺] / [C₁₀H₁₅NO].<span>
Kb = (</span>0,00213 M)² / 0,03287 M = 1,38·10⁻⁴.

3 0

3 years ago

At equilibrium, Keq for the following reaction is equal to 3.5: SO2 + NO20 È SO3 + NO a) Is a beaker containing 2.5 M SO2, 2.0 M

liraira [26]

Keq=[SO3][SO2] /[SO2][NO2]

4 0

3 years ago

Who is good at CHEMISTRY?????????/I NEED HELP

Pavlova-9 [17]

Answer:

With what?

Explanation:

4 0

3 years ago

Read 2 more answers

Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−

kolbaska11 [484]

Answer:

r = 3.61x $10^{-6}$ M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k. $[S2O2^{-8} ]^{x} x [I^{-} ]^{y}$

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

$\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}$

$\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y$

$0.67 = 0.67^x$

x = 1

Now, to find the coefficient y let's do the same for the experiments 1 and 3:

$\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}$

$\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y$

$0.33 = 0.5x 0.67^y$

$0.67 = 0.67^y$

y = 1

Now, we need to calculate the constant k in whatever experiment. Using the first :

$2.6x10^{-6} = kx0.018x0.036$ $kx6.48x10^{-4} = 2.6x10^{-6}$

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = $4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}$

r = 3.61x $10^{-6}$ M/s

7 0

3 years ago