Molecule of dissimilar elements with a net charge remaining is called

The reason that elements in the same group all have similar chemical properties is because they all have the same

sveta [45]

D-number of electrons in the outer energy level.

4 0

3 years ago

A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to

melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the Henderson-Hasselbach (H-H) equation:

pH = pKa + log $\frac{[CH_3COO^-]}{[CH_3COOH]}$

For acetic acid, pKa = 4.75.

We calculate the original number of moles for acetic acid and acetate, using the given concentrations and volume:

CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻

CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to calculate the new pH, by using the new concentrations:

pH = 4.75 + log $\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L}$ = 5.54

6 0

3 years ago

33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b

Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water = 18.88 g

Total mass of the solution = Mass of fructose + Mass of water = M

M = 33.56 g + 18.88 g =52.44 g

Volume of the solution = V = 40.00 mL

Density = $\frac{Mass}{Volume}$

a) Density of the solution:

$\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL$

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = $n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol$

Molar mass of water = 18.02 g/mol

Moles of water= $n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol$

Mole fraction of fructose in this solution: $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}$

$\chi_1=0.1510$

Mole fraction of water = $\chi_2=1-\chi_1=0.8490$

c) Average molar mass of of the solution:

= $\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol$

$=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol$

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

$\frac{\text{Average molar mass}}{\text{Density of the mass}}$

$=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol$

5 0

3 years ago

Ampicillin is a bacteriostatic antibiotic. (Instead of killing bacteria, it inhibits their growth.) Ramon plated bacteria transf

love history [14]

Answer:

A) That resistance in bacteria is produced due to inactivation of ampicillin by the beta lactamase enzyme. This enzyme is expressed by the bla gene found in the plasmid. This enzyme is secreted into the culture medium, thereby inactivating ampicillin. Thanks to this inactivation, the bacteria colonies will be able to develop. After a day of incubation, only those bacteria that took the plasmid that gives them resistance to ampicillin will grow after transformation. After prolonged incubation, two types of colonies can be observed in the culture medium. One large colony with ampicillin resistance, and another small colony, both of which are sensitive to ampicillin.

B) Large colonies are characterized by being resistant to ampicillin. When Ramón isolates the plasmid, he will have the gene that provides resistance to antibiotics. Said plasmid can be used again on those bacteria that are sensitive to ampicillin.

On the other hand, satellite colonies are sensitive to ampicillin. These types of colonies do not have the plasmid that contains the gene that gives ampicillin resistance. It is not possible to isolate any plasmids from these satellite colonies. These satellite bacteria will not be able to grow if they are transferred to a plate containing fresh ampicillin, while large colonies, which possess the plasmid that gives them resistance to ampicillin, will be able to grow on that plate.

Explanation:

3 0

3 years ago

Which organism has a nerve net? A fish B hydra C flatworm D sponge

Kaylis [27]

B. Hydra, because Hydraare an ideal class of Cnidaria to research and on which to run tests.

5 0

3 years ago

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