All categories

3 years ago

Planck’s constant is approximately equal to which of the following?

1 answer:

7 0

Henlo!
Bohr's model was unable to calculate or it required precise information about position of an electron and its velocity. It is very difficult to calculate velocity and position of an electron at the same time because electron i too small to see and may only be observed if peturbed, for example we could hit the electron with another particle such as photon or an electron, or we could apply electric or magnetic field to the electron. This will inevitably change the position of the elctron or its velocity and direction. Heisenberg aid that more precisely we can define the position of an electron, the less certainity we are able to define its velocity and vice versa.
In short, first option is correct one

You might be interested in

damaskus [11]

Mostly in my opinion I think it’s the secound one

8 0

3 years ago

One mole of ANY element contains the:

VMariaS [17]

Answer:

1)The molar mass of an atom is simply the mass of one mole of identical atoms. However, most of the chemical elements are found on earth not as one isotope but as a mixture of isotopes, so the atoms of one element do not all have the same mass.

2)Equally important is the fact that one mole of a substance has a mass in grams numerically equal to the formula weight of that substance. Thus, one mole of an element has a mass in grams equal to the atomic weight of that element and contains 6.02 X 1023 atoms of the element.

3 0

3 years ago

Read 2 more answers

Based on the results you observed for the iodine test and Benedict’s test, is it better to detect enzyme activity by measuring t

dlinn [17]

Answer:

Is better use the Benedict's test by the increase in the amount of the products if the enzyme is a reductase

Explanation:

The Benedict's test works by the reaction of the reducing sugars with the ion cupric of the reactive. If the enzyme is a reductase (degrades polysaccharides into bi o monosaccharides), it should cut the polysaccharide bond and the products would react with the Benedict's cupric ion

I hope you undestand me

3 0

3 years ago

2-butanone is converted into 3-methyl-3-hexanol using a grignard reagent prepared from 1-bromopropane and magnesium metal in thf

Andre45 [30]

Answer:

Here's what I get.

Explanation:

At the end of the reaction you will have a solution of the alcohol in THF.

The microdistillation procedure will vary, depending on the specific apparatus you are using, but here is a typical procedure.

Transfer the solution to a conical vial.
Add a boiling stone.
Attach a Hickman head (shown below) and condenser.
Place the assembly in in the appropriate hole of an aluminium block on top of a hotplate stirrer.
Begin stirring and heating at a low level so the THF (bp 63 °C) can distill slowly.
Use a Pasteur pipet to withdraw the THF as needed.
When all the THF has been removed, raise the temperature of the Al block and distill the alcohol (bp 143 °C).

7 0

4 years ago

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .

5 0

3 years ago