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3 years ago

What is the [OH-] in a solution with a poh of 3.90

Chemistry

2 answers:

zheka24 [161]3 years ago

8 0

Answer:

7.9433E-3

Explanation:

POH=-log OH

3.90=POH

OH=ANTILOG OF -3.90

=7.9433×10^-3

andrew-mc [135]3 years ago

5 0

Answer: The hydroxide ion concentration is $1.26\times 10^{-4}M$

Explanation:

pOH is defined as the negative logarithm of hydroxide ion concentration in the solution.

To calculate the pOH of the reaction, we use the equation:

$pOH=-\log[OH^-]$

where,

$pOH=3.90$

Putting values in above equation, we get:

$3.90=-\log[OH^-]$

$[OH^-]=1.26\times 10^{-4}M$

Hence, the hydroxide ion concentration is $1.26\times 10^{-4}M$

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A 2.50 g sample of solid sodium hydroxide is added to 55.0 mL of 25 °C water in a foam cup (insulated from the environment) and

zlopas [31]

Answer:

37.1°C.

Explanation:

Firstly, we need to calculate the amount of heat (Q) released through this reaction:

∵ ΔHsoln = Q/n

no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.

The negative sign of ΔHsoln indicates that the reaction is exothermic.

∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.

We can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat released to water (Q = 2781.87 J).

m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).

∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)

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2 years ago

Which of the following is an empirical formula? OP205 O C2H602 ON3F6 OC8H18

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Answer:

None are empirical formulas

Explanation:

All are actual compounds. An example of an empirical formula could be CH2O, the empirical formula for carbohydrates like glucose (C6H12O6).

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2 years ago

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3 years ago

You are given a solution that is 518 mM lactose. You need to make up 4.5 L of 16.7 mM solution. What volume do you need to trans

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Answer:

The volume you need to transfer from the stock solution is 0.145 l

Explanation:

Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:

number of moles in volume to transfer = number of moles in the final solution

Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:

Ci * Vi = Cf * Vf

where:

Ci = concentration of the stock solution.

Vi = volume of the stock solution to be transferred.

Cf = concentration of the final solution

Vf = volume of the final solution

Then, replacing with the data:

518 mM * Vi = 16.7 mM * 4.5 l

Vi = 16.7 mM * 4.5 l / 518 mM

Vi = 0.145 l or 145 ml

Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.

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