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3 years ago

A solution has a pH of 2.5. Answer the following questions. pOH = [H]= [OH]=

Chemistry

1 answer:

Sloan [31]3 years ago

4 0

Answer:

pOH = 11.5

[H⁺] = 0.003 M

[OH⁻] = 3 × 10⁻¹² M

Explanation:

The computation is shown below:

Given that

pH = 2.5

Based on the above information

We know that

pH + pOH = 14 ⇒ pOH = 14 - pH

pOH = 14 - 2.5

pOH = 11.5

[H⁺] = 10^(-pH) = 10^(-2.5)

[H⁺] = 0.003 M

[OH⁻] = 10^(-pOH)

= 10^(-11.5)

= 3 × 10⁻¹² M

[OH⁻] = 3 × 10⁻¹² M

Hence, the above represents the answer

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Perform unit conversions and determine Re for the case when a fluid with density of 92.8 lbm/ft3 and viscosity of 4.1 cP (centip

erastovalidia [21]

Answer:

Re=309926.13

Explanation:

density=92.8lbm/ft3*(0.45kg/1lbm)*(1ft3/0.028m3)=1491.43kg/m3

viscosity=4.1cP*((1*10-3kg/m*s)/1cP)=0.0041kg/m*s

velocity=237ft/min*(1min/60s)*(0.3048m/1ft)=1.2m/s

diameter=28inch*(0.0254m/1inch)=0.71m

Re=(density*velocity*diameter)/viscosity=(1491.43kg/m3*1.2m/s*0.71m)/0.0041kg/m*s

Re=309926.13

4 0

3 years ago

What is the ph of a 0.45 m solution of aniline (c6h5nh2)? (pkb  9.40)?

mote1985 [20]

Answer is: pH of aniline is 9.13.
Chemical reaction: C₆H₅NH₂(aq)+ H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).

pKb(C₆H₅NH₂) = 9.40.

Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.

c₀(C₆H₅NH₂) = 0.45 M.

c(C₆H₅NH₃⁺) = c(OH⁻) = x.

c(C₆H₅NH₂) = 0.45 M - x.

Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).

4·10⁻¹⁰ = x² / (0.45 M - x).

Solve quadratic equation: x = c(OH⁻) = 0.0000134 M.

pOH = -log(0.0000134 M.) = 4.87.

pH = 14 - 4.87 = 9.13.

7 0

3 years ago

calculate how many milliliters of 0.142 M NaOH are needed to completely neutralize 21.4 mL of 0.294 M H2C4H4O6.

Flura [38]

Answer:

The answer to your question is 88.7 ml

Explanation:

Data

Volume = ?

Concentration of NaOH = 0.142 M

Volume of H₂C₄H₄O₆ = 21.4 ml

Concentration of H₂C₄H₄O₆ = 0.294 M

Balanced chemical reaction

2 NaOH + H₂C₄H₄O₆ ⇒ Na₂C₄H₄O₆ + 2H₂O

1.- Calculate the moles of H₂C₄H₄O₆

Molarity = moles/volume

Solve for moles

moles = Molarity x volume

Substitution

moles = 0.294 x 21.4/1000

Result

moles = 0.0063

2.- Use proportions to calculate the moles of NaOH

2 moles of NaOH ------------------ 1 moles of H₂C₄H₄O₆

x ------------------ 0.0063 moles

x = (0.0063 x 2) / 1

x = 0.0126 moles of NaOH

3.- Calculate the volume of NaOH

Molarity = moles / volume

Solve for volume

Volume = moles/Molarity

Substitution

Volume = 0.0126/0.142

Result

Volume = 0.088 L or 88.7 ml

3 0

3 years ago

Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined

Furkat [3]

Answer:

The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

$Concentration=\frac{number of moles of solute}{Volume}$

You know the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

$Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}$

The concentrations are:

[PCl₃]= $\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}$

[Cl₂]= $\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}$

[PCl₅]= $\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}$

Replacing:

$Q=\frac{0.08*0.16}{0.04}$

Solving:

Q= 0.32

The reaction quotient (Q) before the reaction is 0.32

4 0

2 years ago

25 grams of radon-222 remains after 15.28 days. How much radon was in the original sample? Write an equation for the decay of ra

Crazy boy [7]

Answer:

half-life = 3.8 days

total time of decay = 15.2 days

initial amount = 100. g

number of half-lives past: 15.2/3.8 = 4 half-lives

4 half-lives = 1/16 remains

100. g x 1/16 = 6.25 g

8 0

2 years ago