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o-na [289]
3 years ago
10

A 35 L tank of oxygen is at 315 K with an internal pressure of 190 atmospheres. How

Chemistry
1 answer:
nata0808 [166]3 years ago
6 0

Answer:

600.7 moles

Explanation:

Applying,

PV = nRT................... Equation 1

Where P = Pressure of oxygen, V = Volume of oxygen, n = number of moles, R = molar gas constant, T = Temperature.

make n the subject of the equation

n = PV/RT............... Equation 2

From the question,

Given: P = 190 atm, V = 35 L, T = 135 K

Constant: R = 0.082 atm.dm³/K.mol

Substitute these values into equation 2

n = (190×35)/(135×0.082)

n = 600.7 moles of xygen

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A city in Laguna generates 0.96 kg per capita per day of Municipal Solid Waste (MSW). Makati City in Metro Manila generates 1.9
mamaluj [8]

Answer:

i) amount of MSW generated:

Laguna: 19200 Kg/day

Makati: 38000 Kg/day

ii) number of trucks to collect twice weekly:

Laguna: 3 trucks

Makati: 5 trucks

iii) volume of MSW in tons that enter landfill/week:

Laguna: 147.84 ton/week

Makati: 292 ton/week

Explanation:

i) Laguna: 0.96 Kg person / day of MSW * 20000 = 19200 Kg MSW / day

⇒ Laguna: 19200 Kg/day * ( 7day/ week ) = 134400 Kg/week

Makati: 1.9 Kg person / day of MSW * 20000 = 38000 Kg MSW / day

⇒ Makati: 38000 Kg/day * ( 7day/week ) = 266000 Kg/week

ii)  truck capacity = 4.4 ton * ( Kg / 0.0011 ton ) = 4000 Kg

⇒ quote/day = 4000 Kg * 0.75 = 3000 Kg

⇒ loads/day = 2 * 3000 kg = 6000 Kg

⇒ operate/week = 5 * 6000 Kg = 30000 Kg

∴ Laguna:  number of trucks needed/week= 134400 / 30000  = 4.48 ≅ 5 trucks

⇒ number of trucks to collect twice weekly = 5 / 2 = 2.5 ≅ 3 trucks

∴ Makati : number of trucks needed/week = 266000 / 30000 = 8.86 ≅ 9 trucks

⇒ number of trucks to collect twice weekly = 9 / 2 = 4.5 ≅ 5 trucks  

iii) enter landfill/week:

Laguna: 134400Kg MSW/week * ( 0.0011 ton/Kg ) = 147.84 ton/week

Makati: 266000Kg MSW/week * ( 0.0011 ton/Kg ) = 292 ton/week

4 0
2 years ago
A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

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Determine the type of alcohol corresponding to each given description or name. 1-pentanol 3-ethyl-3-pentanol 2-hexanol An alcoho
pochemuha

Answer:

1). 1-pentanol - <u>Primary</u>

2). 3-ethyl-3-pentanol - <u>Tertiary</u>

3). 2-hexanol - <u>Secondary</u>

4). Alcohol with two other carbons attached to the carbon with the hydroxyl group - <u>Secondary</u>

5). Alcohol with one other carbon attached to the carbon with the hydroxyl group - <u>Primary</u>

6). Alcohol with three other carbons attached to the carbon with the hydroxyl group - <u>Tertiary</u>

Explanation:

The distinct types of alcohol have been matched with the categories above as per their descriptions provided. In chemistry, alcohols have been categorized into three different categories namely primary, secondary, and tertiary.

In the primary type, those alcohols are involved in which there is an association of hydroxyl group to a primary atom of carbon along with a minimum of two atoms of hydrogen. Example; ethanol.

In the secondary type, the alcohols have an association of carbon atoms to hydroxyl with a single atom of hydrogen and has a formula of '-CHROH.' Example: 2 - propanol.

In the tertiary alcohols, here the association is between the hydroxyl group with the carbon atom that is saturated and possessing 3 atoms of carbon associated with it. It has a formula of '-CR2OH.' Example:  3-ethyl-3-pentanol, -tert -butyl alcohol, etc.

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