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3 years ago

3. Write DISSOCIATION or IONIZATION equations for the following chemicals:

Chemistry

1 answer:

Romashka [77]2 years ago

8 0

Answer:

a) HNO3 -> H+ + NO3- disassociation of Nitric Acid; to yield a Nitrate ion and a Proton, H+, or as a Hydronium ion H3O+

b) H2S04 -> Disassociation of Sulfuric Acid; simple way- 2H+ + SO4- -

c) H2S hydrogen sulphide in water is an acid; thus H+ HS- disassociation.

d) NaOH -> dissociation of Na+ + OH-; this is complete; sodium hydroxide is deliquescent, meaning it will draw water - EVEN from the air! Strong Base

e) Na2CO3 -> 2Na+ CO3- - Ionization of sodium carbonate - a salt

f) Na2S04 -> 2Na+ + SO4 - - ionization of sodium sulphate - a salt

g) NaCl -> Na+ + Cl- ionization of the salt, Sodium Chloride

Explanation:

Salts ionize at different rates; acids or bases dissociate; these are mostly strong acids and NaOH, a strong base.

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Explanation:

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3 years ago

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3 years ago

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2 years ago

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You wish to make a 0.285 M hydroiodic acid solution from a stock solution of 12.0 M hydroiodic acid. How much concentrated acid

Inessa05 [86]

The amount, in mL, of the concentrated acid required, would be 1.1875 mL

<h3>Dilution</h3>

From the dilution equation:

m1v1=m2v2 where m1 and m2 = molarity before and after dilution, and v1 and v2 = volume before and after dilution.

m2 = 0.285M, m1 = 12.0M v2 = 50.0 mL

v1 = m2v2/m1 = 0.285x50/12 = 1.1875 mL

Thus, 1.1875 mL of the acid would be taken and diluted with water up to the 50 mL mark.

More on dilution can be found here: brainly.com/question/13949222

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2 years ago

At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank

hodyreva [135]

Answer:

$\large \boxed{\text{-92 $^{\circ}$C}}$

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

$\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}$

2. Temperature in degrees Celsius

$\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}$

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3 years ago