The proportion of dissolved substances in seawater is usually expressed in

Which compound will give positive Tollen's test?

Andrew [12]

Tollen's test is an organic test that determines whether the sample is an adehyde or not. When aldehydes are present, the reaction forms a silver compound which refers to the other name of Tollen's as silver mirror test. The aldehyde among the choices is C. pentanal

5 0

2 years ago

How many protons does silicon have? a. 2 b. 14 c. 28 d. 28.08

EleoNora [17]

The answers is, B. 14

5 0

2 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago

Major species present when fructose is dissolved in water

Katyanochek1 [597]

The fructose chemical formula is C6H12O6. The answer to the question above regarding the major species present when fructose is dissolved in water (H2O) is "None". No ions are present. It is false that when sugar is dissolved in water there will be strong electrolytes.

7 0

3 years ago

Read 2 more answers

What is the [H+] of a solution with a pH of 6.6? Use a scientific calculator.

svetoff [14.1K]

8.3 × 106 - trust me, it's actually right. You can use the calculator to see if I'm correct. Punch in 8.3 × 106 = 6.6

7 0

3 years ago

Read 2 more answers