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Blababa [14]
3 years ago
8

Calculate the pH of each solution at 25∘C

Chemistry
1 answer:
JulijaS [17]3 years ago
8 0

Answer: pH of HCl =5, HNO3 = 1,

NaOH = 9, KOH = 12

Explanation:

pH = -log [H+ ]

1. 1.0 x 10^-5 M HCl

pH = - log (1.0 x 10^-5)

= 5 - log 1 = 5

2. 0.1 M HNO3

pH = - log (1.0 x 10 ^ -1)

pH = 1 - log 1 = 1

3. 1.0 x 10^-5 NaOH

pOH = - log (1.0 x 10^-5)

pOH = 5 - log 1 = 5

pH + pOH = 14

Therefore , pH = 14 - 5 = 9

4. 0.01 M KOH

pOH = - log ( 1.0 x 10^ -2)

= 2 - log 1 = 2

pH + pOH = 14

Therefore, pH = 14 - 2 = 12

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Molar mass of KOH= 56.1056

( \frac{1 mol}{56.1056 grams})  (\frac{.325 mol}{x grams}) = 18.234 grams
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3 years ago
The sun's inner core is the hottest part of the sun<br><br> true or false
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Read 2 more answers
What is the limiting reactant in a reaction where 10.0 mol of iron is treated with 12.0 mol of bromine? The product that forms i
hammer [34]

<u>Answer:</u> The limiting reagent in the reaction is bromine.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

Given values:

Moles of iron = 10.0 moles

Moles of bromine = 12.0 moles

The chemical equation for the reaction of iron and bromine follows:

2Fe+3Br_2\rightarrow 2FeBr_3

By the stoichiometry of the reaction:

If 3 moles of bromine reacts with 2 moles of iron

So, 12.0 moles of bromine will react with = \frac{2}{3}\times 12.0=8moles of iron

As the given amount of iron is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Hence, bromine is considered a limiting reagent because it limits the formation of the product.

Thus, the limiting reagent in the reaction is bromine.

3 0
3 years ago
Calculate the density of a solid in g/cm3 if it weighs 38.3 kg and has a volume of 0.00463 m3.
RideAnS [48]

Answer: D=8.27 g/cm³

Explanation:

Density is mass/volume. Mass is in grams and volume is in liters. In this case, the problem wants our volume to be in cm³. All we need to do is to make some conversions to convert kg/m³ to g/cm³.

D=\frac{38.3kg}{0.00463m^3} *\frac{(1m)^3}{(100cm)^3} *\frac{1000g}{1kg}

With this equation, the m³ and kg cancel out, and we are left with g/cm³.

D=8.27 g/cm³

3 0
3 years ago
You see the middle layer of the sun’s atmosphere, the _____________, at the start and end of a total eclipse.
Helen [10]

Answer:

Chromosphere

Explanation:

You see the middle layer of the sun’s atmosphere, the Chromosphere, at the start and end of a total eclipse.

4 0
3 years ago
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