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3 years ago

In CaSO4, the oxidation number of Ca is that of S is and that of O is

Chemistry

1 answer:

Assoli18 [71]3 years ago

7 0

Answer: In CaF2, the oxidation number of Ca is +2

, and that of F is -1

. In H2SO4, the oxidation number of H is +1

, that of S is +6

, and that of O is -2

. In CaSO4, the oxidation number of Ca is +2

, that of S is +6

, and that of O is -2

. In HF, the oxidation number of H is +1

, and that of F is -1

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Calculate the [H+] and pH of a 0.000185 M acetic acid solution. Keep in mind that the Ka of acetic acid is 1.76×10−5. This is a

lisov135 [29]

Answer:

[H⁺] = 0.000048936M

pH = 4.31

Explanation:

Acetic acid, CH₃COOH, dissociates in water as follows:

CH₃COOH(aq) ⇄ H⁺(aq) + CH₃COO⁻(aq)

And Ka is defined as:

Ka = 1.76x10⁻⁵ = [H⁺] [CH₃COO⁻] / [CH₃COOH]

<em>Where [] are equilibrium concentrations of the species.</em>

<em />

The 0.000185M of acetic acid will decreases X, and X of [H⁺] and [CH₃COO⁻] will be produced. That means Ka is:

1.76x10⁻⁵ = [X] [X] / [0.000185 - X]

3.256x10⁻⁹ - 1.76x10⁻⁵X = X²

3.256x10⁻⁹ - 1.76x10⁻⁵X - X² = 0

Solving for X:

X = -0.000066M → False solution. There is no negative concentrations.

X = 0.000048936

As [H⁺] = X,

[H⁺] = 0.000048936M

And pH = -log [H⁺]

8 0

3 years ago

How many molecules are there in 5.9 moles of NaCl?

choli [55]

The number of molecules= 3.55 x 10²⁴

<h3>Further explanation</h3>

Given

5.9 moles of NaCl

Required

The number of molecules

Solution

The mole is the number of particles(molecules, atoms, ions) contained in a substance

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

So for 5.9 moles :

= 5.9 x 6.02 x 10²³

= 3.55 x 10²⁴

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3 years ago

What happens when you add or change a subscript

zavuch27 [327]

When you add or change the subscript, you are changng the number of atoms in the substance.

7 0

4 years ago

In order to be sure that your results are accurate, you must be able to ______ your experiment.

mina [271]

The answer is A, you must repeat your experiment to ensure you results are accurate. That is the best way to know for sure and it is what you must do to every experiment.

7 0

3 years ago

PLEASE HELP FAST IM ABOUT TO FAIL PLEASE HELP PLEASE HELP FAST HELP HELP THIS IS DESPRATE

MariettaO [177]

1. 0.240 liters of water would be needed to dissolve 21.6 g of lithium nitrate to make a 1.3 M (molar) solution.

2. 2.9 M is the molarity of a solution made of 215.1 g of HCl is dissolved to make 2.0 L of solution.

3.83.3 ml of concentrated 18 M H2SO4 is needed to prepare 250.0 mL of a 6.0 M solution.

4. 135 ml of stock HBr will be required to dilute the solution.

5. 150 ml of water should be added to 50.0 mL of 12 M hydrochloric acid to make a 4.0 M solution

6. The pH of the resulting solution is 13.89

Explanation:

The formula used in solving the problems is

number of moles= $\frac{mass}{atomic mass of one mole}$ 1st equation

molarity = $\frac{number of moles}{volume}$ 2nd equation

Dilution formula

M1V1 = M2V2 3rd equation

1. Data given

mass of Lithium nitrate = 21.6 grams

atomic mass of on emole lithium nitrate = 68.946 gram/mole

Molarity is given as 1.3 M

VOLUME=?

Calculate the number of moles using equation 1

n = $\frac{21.6}{68.946}$

= 0.313 moles of lithium nitrate.

volume is calculated by applying equation 2.

volume = $\frac{0.313}{1.3}$

= 0.240 litres of water will be used.

2. Data given:

mass of HCl = 215.1 gram

atomic mass of HCl = 36.46 gram/mole

volume = 2 litres

molarity = ?

using equation 1 number of moles calculated

number of moles = $\frac{215.1}{36.46}$

number of moles of HCl = 5.899 moles

molarity is calculated by using equation 2

M = $\frac{5.899}{2}$

= 2.9 M is the molarity of the solution of 2 litre HCl.

3. data given:

molarity of H2SO4 = 18 M

Solution to be made 250 ml of 6 M

USING EQUATION 3

18 x V1= 250 x 6

V1 = 83.3 ml of concentrated 18 M H2SO4 will be required.

4. data given:

M1= 10M, V1 =?, M2= 3 ,V2= 450 ml

applying the equation 3

10 x VI = 3x 450

V1 = 135 ml of stock HBr will be required.

5. Data given:

V1 = 50 ml

M1= 12 M

V2=?

M2= 4

applying the equation 3

50 x 12 = 4 x v2

V2 = 150 ml.

6. data given:

HCl + NaOH ⇒ NaCl + H20

molarity of NaOH = 0.525 M

volume of NaOH = 25 ml

molarity of acid HCl= 75 ml

volume of HCl = 0.335 ml

pH=?

Number of moles of NaOH and HCl is calculated by using equation 1 and converting volume in litres

moles of NaOH = 0.0131

moles of HCl= 0.025 moles

The ratio of moles is 1:1 . To find the unreacted moles of acid and base which does not participated in neutralization so the difference of number of moles of acid minus number of moles of base is taken.

difference of moles = 0.0119 moles ( NaOH moles is more)

Molarity can be calculated by using equation 1 in (25 +75 ml) litre of solution

molarity = $\frac{0.0119}{0.1}$

= 0.11 M (pOH Concentration)

14 = pH + pOH

pH = 14 - 0.11

pH = 13.89

3 0

3 years ago