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Otrada [13]
3 years ago
13

In this experiment you are directed to add a limited amount of NaOH (aq) and then an excess amount. A similar strategy will be u

sed with the addition of NH3(aq). Why? Group of answer choices Both NaOH and NH3 are weak bases. Both NaOH and NH3 are strong bases. Initially a metal hydroxide may form which becomes soluble when more base is added and the metal complex forms. Insoluble metal complexes are first formed which then are replaced by insoluble metal hydroxide precipitates. Initially a complex ion may form which becomes insoluble when more base is added and the metal hydroxide precipitate forms.
Chemistry
1 answer:
Verizon [17]3 years ago
7 0

The reason for adding a limited amount and then an excess amount is that initially a metal hydroxide may form which becomes soluble when more base is added and the metal complex forms.

In qualitative analysis is a common to add the base in drops and then in excess. When added in drops, the metal hydroxide is formed. This metal hydroxide is often insoluble.

After this metal hydroxide is formed, the base could be added in excess such that the metal hydroxide dissolves in the excess base by forming a complex.

For instance;

CuCl2(aq) + 2NaOH(aq)  -------> Cu(OH)2(s) + 2NaCl(aq)

Cu(OH)2(s) + 2OH^-(aq) -------> [Cu(OH)4]^2+(aq)

Learn more: brainly.com/question/1527403

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How much energy is required to convert 15.0 g of ice at −106 °C to water vapor at 125 °C? Specific heats are 2.09 J/g K for both
myrzilka [38]

Answer:

49.3 kJ of energy is required

Explanation:

An exercise of calorimetry at its best

First of all, convert the ice to water before melting.

Q = ice mass . C . ΔT

Q = 15 g . 2.09 J/g°C (0° - (-106°C)

15 g . 2.09 J/g°C . 106°C = 3323.1 J

Now we have to melt the ice, to change its state

Q = mass . latent heat of fusion

Q = 15 g . 0.335 kJ/g = 5.025 kJ .1000 = 5025 J

After that, we have liquid water at 0° and the ice has melted completely. We have to release energy to make a temperature change, to 100° (vaporization)

Q = 15g . 4.18 J/g°C (100°C - 0°C)

Q = 6270 J

Water has been vaporizated so we have to calculate, the state change.

Q = mass . latent heat of vap

Q = 15 g. 2.260 kJ/g

Q = 33.9 kJ (.1000) = 33900 J

Finally we have to increase temperature from 100°C to 125°C

Q = 15 g . 2.09 J/g°C . (125°C - 100°C)

Q = 783.75 J

To know how much energy is required to conver 15 g of ice, to water vapor at 125°C, just sum all the heat released.

3323.1 J + 5025 J + 6270 J + 33900 J + 783.75 J = 49301.85 joules.

Notice I have to convert kJ to J in two calcules to make the sum.

49301.85 joules / 1000 = 49.3 kJ

4 0
3 years ago
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When selling on the street, dealers may not know the purity of the ketamine they have, and thus users do not know exactly how mu
fiasKO [112]

Answer:

a. 6.15 mL b. 30.73 mL

Explanation:

a. What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg?

Since we have 0.250 g of ketamine in 1/4 cup of water and 1 cup of water equals 236.5 mL, we need to find the concentration of ketamine we have.

So concentration of ketamine C = mass of ketamine, m/volume of water, V

m = 0.250 g and V = 1/4 cup = 1/4 × 236.5 mL = 59.125 mL

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Since the user has a mass of 65 kg and requires a high at 0.400 mg/kg, the mass of ketamine for this high is M = 65 kg × 0.400 mg/kg = 26 mg

Since mass, M = concentration ,C × volume, V

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b. What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?

Since the concentration of ketamine is C = 4.23 mg/mL, and Since the user has a mass of 65 kg and requires an injection of 2.00 mg/kg to be unconscious, the mass of ketamine required to be unconscious is M' = 65 kg × 2.00 mg/kg = 130 mg

Since mass, M' = concentration ,C × volume, V

M' = CV

V = M/C

The volume of ketamine required for the 2.00 mg/kg unconscious injection is

V = 130 mg/4.23 mg/mL

V = 30.73 mL

5 0
3 years ago
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