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2 years ago

What is the pOH of a solution that has a OH - concentration equal to 1.3x 10^-10

Chemistry

2 answers:

Paul [167]2 years ago

7 0

Answer: 9.88 (approx10)

Explanation:

The pOH of a solution is the negative logarithm of the hydroxide-ion concentration

larisa [96]2 years ago

3 0

Answer: pOH = 9.88

Explanation:pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

Acids have pH ranging from 1 to 6.9, bases have pH ranging from 7.1 to 14 and neutral solutions have pH equal to 7.

$pH=-\log [H^+]$

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

Putting in the values:

$pOH=-\log[1.3\times 10^{-10}]$

$pOH=9.88$

$pH+pOH=9.88$

$pH=14-9.88=4.12$

Thus as pH is less than 7 the solution is acidic and pOH is 9.88.

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For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow

lora16 [44]

Answer:

Rate constant of the reaction is $3.3\times 10^{-3} M^{-2} s^{-1}$ .

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

$R=k[A]^a[B]^b[C]^c$

Rate(R) of the reaction in trail 1 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.30 M$

$R=9.0\times 10^{-5} M/s$

$9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c$ ...[1]

Rate(R) of the reaction in trail 2 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.90 M$

$R=2.7\times 10^{-4} M/s$

$2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c$ ...[2]

Rate(R) of the reaction in trail 3 ,when :

$[A]=0.60 M,[B]=0.30 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c$ ...[3]

Rate(R) of the reaction in trail 4 ,when :

$[A]=0.60 M,[B]=0.60 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c$ ...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

$R=k[A]^2[B]^0[C]^1$

Rate constant of the reaction = k

$9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1$

$k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}$

$k=3.3\times 10^{-3} M^{-2} s^{-1}$

7 0

3 years ago

A student needs to prepare 100. mL of 0.612 M Cu(NO3)2 solution. What mass, in grams, of copper(II) nitrate should the student u

Temka [501]

Answer: 11.5 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution

$Molarity=\frac{n\times 1000}{V_s}$

where,

Morality = 0.612 M

n= moles of solute

$V_s$ = volume of solution in ml = 100 ml

Now put all the given values in the formula of molarity, we get

$0.612=\frac{n\times 1000}{100ml}$

$n=0.0612moles$

$Mass={\text {moles of solute }}{\times {\text {molar mass}}=0.0612moles\times 187.56g/mol=11.5g$

Therefore, the mass of copper (II)nitrate required is 11.5 grams

3 0

2 years ago

How many oxygen is produced in Laboratory?

xxTIMURxx [149]

Answer:

Explanation:

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In this lab, oxygen will be generated as a product of the decomposition of hydrogen peroxide. A catalyst is used to speed up the rate of the decomposition reaction, which would otherwise be too slow to use as a source of oxygen. The catalyst does not get consumed by the reaction, and can be collected for re-use once the reaction is complete. The particular catalyst used in this lab is manganese(IV) oxide.

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2 years ago

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KATRIN_1 [288]

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2 years ago

Briefly explain formality in solution.

Zielflug [23.3K]

Answer:

Both molarity and formality express concentration as moles of solute per liter of solution. Formality is a substance's total concentration in solution without regard to its specific chemical form. ... The formality of a solution is defined as the number of formula mass of any solute dissolved in 1 litre of solution.

4 0

2 years ago