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3 years ago

5. Identify the oxidation state of Ca in CaF2(aq). Ca(s) + 2 HF(aq) → CaF2(aq) + H2(g)

Chemistry

1 answer:

NikAS [45]3 years ago

8 0

Answer:

Explanation:

Chemical equation:

Ca + 2HF → CaF₂ + H₂

The oxidation state of Ca on left hand side is 0 while on the right hand side when it form CaF₂ its oxidation state becomes +2 because it loses two. These two electrons are accepted by fluorine and complete its octet while Ca loses its two out most electron to complete the octet.

In this reaction oxidation of calcium takes place and calcium itself is reducing agent.

Reducing agents:

Reducing agents reduced the other element are it self gets oxidized.

Oxidizing agents:

Oxidizing agents oxidize the other elements and itself gets reduced.

Consider another examples.

4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O → CO₂ + H₂

the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.

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A buffer is prepared by adding 300. 0 ml of 2. 0 mnaoh to 500. 0 ml of 2. 0 mch3cooh. what is the ph of this buffer? ka= 1. 8 10

Anton [14]

The Henderson-Hasselbalch equation can be used to determine the pH of the buffer from the pKa value. The pH of the buffer will be 4.75.

<h3>What is the Henderson-Hasselbalch equation?</h3>

Henderson-Hasselbalch equation is used to determine the value of pH of the buffer with the help of the acid disassociation constant.

Given,

Acid disassociation constant (ka) = 1. 8 10⁻⁵

Concentration of NaOH = 2.0 M

Concentration of CH₃COOH = 2.0 M

pKa value is calculated as,

pKa = -log Ka

pKa = - log (1. 8 x 10⁻⁵)

Substituting the value of pKa in the Henderson-Hasselbalch equation as

pH = - log (1. 8 x 10⁻⁵) + log [2.0] ÷ [2.0]

pH = - log (1. 8 x 10⁻⁵) + log [1]

= 4.745 + 0

= 4.75

Therefore, 4.75 is the pH of the buffer.

Learn more about the Henderson-Hasselbalch equation here:

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6 0

1 year ago

At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil

Romashka [77]

Answer:

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 104 kPa

$P_2$ = final pressure of gas = 52 kPa

$V_1$ = initial volume of gas = $2.0m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $21.1^oC=273+21.1=294.1K$

$T_2$ = final temperature of gas = $-5.0^oC=273+(-5.0)=268 K$

Now put all the given values in the above equation, we get:

$\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}$

$V_2=3.64 m^3$

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

3 0

2 years ago

What are the correct coefficients for each reactant and product when this equation is balanced? Make sure to select the correct

Mars2501 [29]

Answer: 44

Explanation:

5 0

3 years ago

Gold has a molar mass of 197 g/mol. (a) how many moles of gold are in a 3.98 g sample of pure gold? (b) how many atoms are in th

Natali5045456 [20]

A.)49.4974874 moles or 49.5 moles
B.)2.980808730172671e+25 or 3e+25

6 0

3 years ago

If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate (KNO3), what is the limiting reagent?

GenaCL600 [577]

K:

m=155g
M=39g/mol

n = 155g / 39g/mol ≈ 3,97mol

KNO₃:

m=122g
M=101g/mol

n = 122g/101g/mol = 1,21mol

2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent

KNO₃ is limiting reagent

5 0

3 years ago