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earnstyle [38]
2 years ago
5

An average human being has about 5.0 L of blood in his or her body. If an average person were to eat 37.7 g of sugar (sucrose, ,

342.30 g/mol), and all that sugar were dissolved into the bloodstream, how would the molarity of the blood sugar change
Chemistry
1 answer:
Viktor [21]2 years ago
6 0

Answer: 0.0220275 M

Explanation:

So, we are given the following data or parameters which are going to help in solving this particular Question/problem.

=> Averagely, we have the volume = 5.0 L of blood in human body .

=> Mass of sugar eaten = 37.7 g of sugar (sucrose, 342.30 g/mol).

Therefore, the molarity of the blood sugar change can be calculated as below:

The molarity of the blood sugar change = (1/ volume) × mass/molar mass.

Thus, the molarity of the blood sugar change = (1/5) × 37.7/342.30 = 0.0220275 M.

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Which subatomic particle do you think is most important to an atom’s identity?
Sonja [21]
The subatomic particle that identifies the atom is the number of protons. This is what distinguishes an element that is is flammmable, hydrogen to one that is essential component in water, oxygen.
5 0
3 years ago
Jada Peterson
barxatty [35]

Answer: A

Explanation:

4 0
2 years ago
Are the following combinations allowed? If not, show two ways to correct them:
mafiozo [28]

The following combination of n=3 ; l=1 ; ml=-2 is not allowed. One way to correct this would be by changing the azimuthal quantum number, l and the other way would be to change the magnetic quantum number, m.

<h3>Is the following combination n=3; l=1; ml=-2 allowed or not.? If not, suggest two ways through which it can be corrected.</h3>

The following combination of n=3 ; l=1 ; ml=-2 is not allowed.

There are several rules that need to be followed for assigning electron quantum numbers. They are:

1. Principal quantum number should be 1 ≤ n

2. Azimuthal quantum number, 0 ≤ l ≤ n − 1

3. Magnetic quantum number, -l ≤ ml ≤ l

4. Spin quantum number as either -1/2 or +1/2

For n = 3,

l should be n - 1 or n - 2 or n - 3 = 2, 1, 0 respectively.

If we choose l = 1 then ml should be -1, 0 and +1

Therefore, one way to correct the combination would be to change the magnetic quantum number to -1

If we choose l = 2 then ml would be -2, -1, 0, +1, +2

Thus, another way to correct the combination is to choose the azimuthal quantum number as 2.

Thus, the following combination of n=3; l=1; ml=-2 is not allowed. One way to correct this would be by changing the azimuthal quantum number, l and the other way would be to change the magnetic quantum number, m.

To learn more about quantum numbers refer:

brainly.com/question/5927165

#SPJ4

5 0
1 year ago
Determine the molar mass for ammonia (NH3) in g/mol.
navik [9.2K]

Answer:

17.031 g/mol

Explanation:

5 0
3 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
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