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dimulka [17.4K]
3 years ago
5

I REALLY NEED HELP can someone help me please I’ll mark you brainliest! :)

Chemistry
1 answer:
Yanka [14]3 years ago
5 0

Answer:

A

Explanation:

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An atom has 6 electrons in its outer orbit. It will most likely form an ion with a charge of:
GenaCL600 [577]

Answer:

c

Explanation:

4 0
3 years ago
Intact cells of two unknown cell types were placed into solutions with different concentrations of NaCl. Type I cells swelled an
Bess [88]

Answer:

Cell Type I: Animal cell surrounded by a plasma membrane only.

Cell Type II: Plant cell surrounded by a plasma membrane and a cell wall.

Explanation:

Hello,

In this case, we must remember that animal cells are covered only by a plasma membrane, which is not enough to keep the cell from bursting as no limit for the inlet salt is established, causing it both to swell and burst. On the other hand, plant cells are covered by both a plasma membrane and a cell wall which contributive effect allow the inlet of the salt but prevent the cell to burst as the cell wall is rigid. In such a way, based on the described situation, one infers that the cell type I is an animal cell surrounded by a plasma membrane only the cell type II is a plant cell surrounded by a plasma membrane and a cell wall.

Best regards.

3 0
3 years ago
Write the equilibrium constant expression for this reaction: 2H+(aq)+CO−23(aq) → H2CO3(aq)
MrRissso [65]

Answer:

Equilibrium constant expression for \rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) \rightleftharpoons H_2CO_3\, (aq):

\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}.

Where

  • a_{\mathrm{H_2CO_3}}, a_{\mathrm{H^{+}}}, and a_{\mathrm{CO_3}^{2-}} denote the activities of the three species, and
  • [\mathrm{H_2CO_3}], \left[\mathrm{H^{+}}\right], and \left[\mathrm{CO_3}^{2-}\right] denote the concentrations of the three species.

Explanation:

<h3>Equilibrium Constant Expression</h3>

The equilibrium constant expression of a (reversible) reaction takes the form a fraction.

Multiply the activity of each product of this reaction to get the numerator.\rm H_2CO_3\; (aq) is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be \left(a_{\mathrm{H_2CO_3\, (aq)}}\right).

Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient "2" on the product side of this reaction. \rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) is equivalent to \rm H^{+}\, (aq) + H^{+}\, (aq) + {CO_3}^{2-}\, (aq). The species \rm H^{+}\, (aq) appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:

\left(a_{\mathrm{H^{+}}}\right)\cdot \left(a_{\mathrm{H^{+}}}\right)\cdot \, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right = \left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right.

That's where the exponent "2" in this equilibrium constant expression came from.

Combine these two parts to obtain the equilibrium constant expression:

\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}.

<h3 /><h3>Equilibrium Constant of Concentration</h3>

In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous "(\rm aq)" species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:

\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}.

8 0
3 years ago
Calculate the molar mass of RbOH
Ghella [55]

Hey there!

RbOH

Rb: 1 x 85.468 = 85.468

O:  1 x 16 = 16

H: 1 x 1.008 = 1.008

------------------------------------

                   102.476

The molar mass of RbOH is 102.476 g/mol.

Hope this helps!

5 0
3 years ago
Heptane and water do not mix, and heptane has a lower density (0.684 g/mL) than water (1.00 g/mL). A graduated cylinder contains
Alekssandra [29.7K]

Answer:

TRY HARD!

Explanation:

DO UR WORK ANG GET IT DONE!

3 0
1 year ago
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