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3 years ago

When the [H+] in a solution is 1.7 × 10−9 M, what is the pOH

2 answers:

7 0

The pOH of a solution is related to the concentration of hydroxide ions in a solution. Also, it is related to the pH of the solution since the sum of pOH and pH is equal to 14. We can use this relation to solve this problem. We do as follows:

pH = -log[1.7 × 10−9 M] = 8.8
pOH + pH = 14
pOH + 8.8 = 14
pOH = 5.2

LuckyWell [14K]3 years ago

6 0

H+ in a solution is 1.7 * 10-9m the answer is below
The pOH is 5.23.

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Based on the activity series, which one of the reactions below will occur (Spontaneous)? A. Mn (s) + NiCl2 (aq) → MnCl2 (aq) + N

kap26 [50]

Answer:

Mn (s) + NiCl2 (aq) → MnCl2 (aq) + Ni

Explanation:

The order of displacement of metals from aqueous solution by another metal is defined by the activity series of metals.

The activity series arranges metals in order of reactivity and increasing electrode potentials. The less negative the electrode potential of a metal is, the less reactive it is and the lower it is found in the activity series.

Nickel has a less negative electrode potential than manganese hence it is displaced from an aqueous solution of its salt by manganese spontaneously.

8 0

3 years ago

Systematic or Random Error? -This type of error affects overall accuracy but does not necessarily affect precision. This type of

love history [14]

Answer:

This type of error affects overall accuracy but does not necessarily affect precision. - Systematic error

This type of error affects precision but does not necessarily affect overall accuracy. - Random error

This type of error occurs if you use a buret that was calibrated incorrectly when it was made. - Systematic error

You can minimize this type of error by taking repeated measurements. - Random error

Explanation:

Systematic errors are errors that are attributable to instrument being used during measurement or consistent incorrect measurement during a research. They are consistently and repeatedly committed during measurements and therefore affect the overall accuracy of measurements. A person committing systematic error can have precise repeated measurement but will be far from being accurate.

Random errors on the other hand has no pattern and are usually unavoidable because they cannot be predicted. When sufficient replicate measurements are made, such errors are reduced to the barest minimum and usually do not affect the overall accuracy of measurements.

3 0

3 years ago

The gas in a closed container has a pressure of 3.00 x 10² kPa 30 ° C. What will the pressure be if the temperature is lowered t

Rainbow [258]

Answer: 100kPa

Explanation:

P1 = 3.00 x 10² kPa , P2 =?

T1 = 30°C = 30 +273 = 303k

T2 = —172°C = —172 + 273 = 101k

P1/T1 = P2/T2

3 x 10² / 303 = P2 / 101

P2 = (3 x 10² / 303) x 101

P2 = 100kPa

6 0

3 years ago

Calculate the amount of gold deposited when a current of 5A is passed through a solution of gold salt for 2hrs 15mins. If the sa

Nataliya [291]

Answer:

a. 82.68 g b. 9.8 min

Explanation:

a. The amount of gold deposited by 5 A current in 2 hrs 15 mins

Since charge Q = It where I = current = 5 A and time = 2 hrs 15 mins = 2 × 60 min + 15 min = 120 + 15 min = 135 min = 135 × 60 s = 8100 s

Q = 5 A × 8100 s

= 40500 C

Also, Q = nF where n = number of moles of gold deposited and F = Faraday's constant = 96500 C

n = Q/F = 40500 C/96500 C = 0.4195 moles ≅ 0.42 mole

Now n = m/M where m = mass of gold and M = molar mass of gold = 197

m = nM

= 0.42 × 197 g

= 82.68 g

b. The time taken for 6g of gold to be deposited.

We first find the number of moles of gold in 6g of gold

Since n = m/M and m = 6 g

n = 6/197 = 0.0305 mole

Q = It = nF

t = nF/I

= 0.0305 mol × 96500 C/5 A

= 2939.09 mol C

= 587.82 s

Changing t to minutes

587.82/60 s = 9.8 min

8 0

3 years ago

Given the following reaction: 2K3PO4 + AL2(CO3)3 = 3K2CO3 + 2ALPO4 If I perform this reaction with 150 g of potassium phosphate

Novosadov [1.4K]

Answer : The theoretical yield of potassium carbonate is, 146.483 g

The percent yield of potassium carbonate is, 85.33 %

Solution : Given,

Mass of $K_3PO_4$ = 150 g

Mass of $Al_2(CO_3)_3$ = 90 g

Molar mass of $K_3PO_4$ = 212.27 g/mole

Molar mass of $Al_2(CO_3)_3$ = 233.99 g/mole

Molar mass of $K_2CO_3$ = 138.205 g/mole

First we have to calculate the moles of $K_3PO_4$ and $Al_2(CO_3)_3$

$\text{ Moles of }K_3PO_4=\frac{\text{ Mass of }K_3PO_4}{\text{ Molar mass of }K_3PO_4}=\frac{150g}{212.27g/mole}=0.7066moles$

$\text{ Moles of }Al_2(CO_3)_3=\frac{\text{ Mass of }Al_2(CO_3)_3}{\text{ Molar mass of }Al_2(CO_3)_3}=\frac{90g}{233.99g/mole}=0.3846moles$

The given balanced reaction is,

$2K_3PO_4+Al_2(CO_3)_3\rightarrow 3K_2CO_3+2AlPO_4$

From the given reaction, we conclude that

2 moles of $K_3PO_4$ react with 1 mole of $Al_2(CO_3)_3$

0.7066 moles of $K_3PO_4$ react with $\frac{1}{2}\times 0.7066=0.3533$ moles of $Al_2(CO_3)_3$

But the moles of $Al_2(CO_3)_3$ is, 0.3846 moles.

So, $Al_2(CO_3)_3$ is an excess reagent and $K_3PO_4$ is a limiting reagent.

Now we have to calculate the moles of $K_2CO_3$ .

As, 2 moles of $K_3PO_4$ react to give 3 moles of $K_2CO_3$

So, 0.7066 moles of $K_3PO_4$ react to give $\frac{3}{2}\times 0.7066=1.0599$ moles of $K_2CO_3$

Now we have to calculate the mass of $K_2CO_3$ .

$\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3\times \text{ Molar mass of }K_2CO_3$

$\text{ Mass of }K_2CO_3=(1.0599moles)\times (138.205g/mole)=146.483g$

The theoretical yield of potassium carbonate = 146.483 g

The experimental yield of potassium carbonate = 125 g

Now we have to calculate the % yield of potassium carbonate.

Formula for percent yield :

$\% yield=\frac{\text{ Theoretical yield}}{\text{ Experimental yield}}\times 100$

$\% \text{ yield of }K_2CO_3=\frac{125g}{146.483g}\times 100=85.33\%$

Therefore, the % yield of potassium carbonate is, 85.33%

6 0

2 years ago