calculate the change in ph when 9.00 ml of 0.100 m hcl(aq) is added to 100.0 ml of a buffer solution that is 0.100 m in nh3(aq) and 0.100 m in nh4cl(aq). consult

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Answer 1

The change in ph when 9.00 ml of 0.100 m hcl(aq) is added to 100.0 ml of a buffer solution that is 0.100 m in nh3(aq) and 0.100 m in nh4cl(aq). consult

1)

Step 1: Calculate initial pH

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {0.1/0.1}

= 4.745

use:

PH = 14 - pOH

= 14 - 4.7447

= 9.2553

Step 2: Calculate pH after adding HCl

mol of HCl added = 0.1M *9.0 mL = 0.9 mmol

NH3 will react with H+ to form NH4+

Before Reaction:

mol of NH3 = 0.1 M *100.0 mL

mol of NH3 = 10 mmol

mol of NH4+ = 0.1 M *100.0 mL

mol of NH4+ = 10 mmol

after reaction,

mol of NH3 = mol present initially - mol added

mmol of NH3 = (10 - 0.9) mmol

mol of NH3 = 9.1 mmol

mol of NH4+ = mol present initially + mol added

mol of NH4+ = (10 + 0.9) mmol

mol of NH4+ = 10.9 mmol

since volume is both in numerator and denominator, we can use mol instead of concentration

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

After the reaction, the following equation holds true: mol of NH3 = mol present initially - mol added mmol of NH3 = (10 - 0.9) mmol mol of NH3 = 9.1 mmol mol of NH4+ = mol present initially + mol added mol of NH4+ = (10 + 0.9) Since volume appears in both the numerator and denominator of the equation, we can use mol in place of concentration: mmol mol of NH4+ = 10.9 mmol

Kb = 1.8*10-5 pKb = -log(Kb) = -log(1.8*10-5) = 4.745

Employ: pOH = pKb + log [conjugate acid]/[base] = 4.745 + log [10.9/9.1] = 4.823

employ: PH = 14 - pOH = 14 - 4.8231 = 9.1769

Step 3: Subtract the final pH from the initial pH to arrive at a change delta pH of -0.0784 (9.1769 - 9.2553).

Incorrect: -0.0784

2) The reaction between NH4+ and OH- will produce NH3 when mol of NaOH supplied is 0.1M *9.0 mL = 0.9 mmol.

NH3 = 0.1 M * 100.0 mL mol of NH3 = 10 mmol NH4+ = 0.1 M * 100.0 mL mol of NH4+ = 10 mmol

After reaction, the equation is: Mol of NH3 = Mol originally present + Mol of NH3 added = (10 + 0.9). mmol mol of NH3 equals 10.9 mmol mol of NH4+ equals mol initially present minus mol added mol of NH4+ = (10 - 0.9) Since volume is in both the numerator and the denominator of mmol mol of NH4+ = 9.1 mmol, we can substitute mol for concentration.

Kb = 1.8*10-5 pKb = -log(Kb) = -log(1.8*10-5) = 4.745

Make use of: pOH = pKb + log [conjugate acid]/[base] = 4.745 + log [9.1/10.9] = 4.666

employ: PH = 14 - pOH = 14 - 4.6663 = 9.3337

delta pH = final pH - initial pH

= 9.3337 - 9.2553

= 0.0784

Answer: 0.0784

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We are considering whether a toxic compound will easily move through the soil and enter the groundwater. The property that must be identified is: D. Soil sorption coefficient (Kd).

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Soil sorption (adsorption) coefficient, in short Kd, measures the amount of chemical substances absorbed into the soil per amount of water. To calculate the Kd value, we need to divide the concentration of the solid (mg/kg dry solid) by the concentration in the pore water (mg/L). Hence, the correct answer is D. Soil sorption coefficient (Kd).

The question seems incomplete. The complete query is as follows:

When considering whether a toxic compound will easily move through soil and enter the groundwater—which of the following properties must be identified?

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c. Octanol-water coefficient (Kow)

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when a cold drink is taken from a refrigerator, its temperature is 5°c. after 25 minutes in a 20°c room its temperature has increased to 10°c. (round your answers to two decimal places.)

Answers

Temperature after 40 minutes is 12.15 degree C

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Setting T=T0 (initial temperature of cold drink) at t=0 then we have

T0= [tex]Ce^{0}[/tex]+ Ts ⟺ C=T0−Ts

Now, we have

T=(T0−Ts)ekt+Ts

ekt=T−TsT0−Ts

t=1kln(T−TsT0−Ts)(1)

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25=1kln(10−205−20)

k=125ln(23)

a) Temperature T after t=40 minutes, setting corresponding values in (1), we get

40=1125ln(23)ln(T−205−20)

ln(20−T15)=4025ln(23)

20−T15=(23)8/5

T=20−15([tex]2/3^{8/5}[/tex]) ≈ 12.15 degree C

Temperature is a unit used to represent how hot or cold something is. It can be stated using the Celsius or Fahrenheit scales, among others. Temperature shows which way heat energy will naturally flow, i.e., from a hotter (body with a higher temperature) to a colder (body with a lower temperature) (one at a lower temperature)

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solid ammonium chloride, nh4cl, is formed by the reaction of gaseous ammonia, nh3, and hydrogen chloride, hcl. nh3(g) hcl(g)⟶nh4cl(s) a 4.98 g sample of nh3 gas and a 4.98 g sample of hcl gas are mixed in a 0.50 l flask at 25 ∘c. identify the limiting reagent. nh3 hcl nh4cl how many grams of nh4cl will be formed by this reaction? mass: g what is the pressure in atmospheres of the gas remaining in the flask? ignore the volume of solid nh4cl produced by the reaction.

Answers

The limiting reagent in this reaction is the hydrogen chloride (HCl).

What is reagent?
A reagent, also known as an analytical reagent, is a substance as well as compound that is added to a system in chemistry to bring about a chemical reaction or check to see if one occurs. Although the terms "reagent" and "reactant" are frequently used interchangeably, "reactant" refers to a substance that is consumed during a chemical reaction. Despite being a part of the reaction mechanism, solvents are not typically referred to as reactants. Catalysts are not reactants because they are not consumed by the reaction. The reactants in biochemistry are frequently referred to as substrates, particularly in relation to enzyme-catalyzed reactions.

The limiting reagent in this reaction is the hydrogen chloride (HCl). This is because it requires more moles of HCl than ammonia to produce ammonium chloride (NH4Cl). The number of moles of HCl is 0.0998, while the number of moles of ammonia is 0.0498, so HCl is the limiting reagent.

The amount of NH4Cl that will be formed will be 4.98 g, since this is the amount of the limiting reagent, HCl.

The pressure of the gas remaining in the flask will be unchanged, since no gas is consumed in the reaction, only solids. Therefore, the pressure will remain at 0.50 atmospheres.

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