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Luba_88 [7]
2 years ago
14

When a nonmetal oxide reacts with water, it forms an oxoacid with the same oxidation number as the nonmetal. Give the name and f

ormula of the oxide used to prepare each of these oxoacids:
(a) hypochlorous acid;
Chemistry
1 answer:
vaieri [72.5K]2 years ago
5 0

Addition of chlorine to water gives both hydrochloric acid (HCl) and hypochlorous acid (HClO)

What are Transition metal oxides ?

Transition metal oxides (TMOs) are another class of nanomaterials, frequently used as anode in alkaline batteries due to their distinctive properties such as abundant active sites, short diffusion pathways, low preparation cost, high theoretical capacity and distinct reaction mechanism.

Cl2 + H2O ⇌ HClO + HCl

Cl2 + 4 OH− ⇌ 2 ClO− + 2 H2O + 2 e−

Cl2 + 2 e− ⇌ 2 Cl−

The acid can also be prepared by dissolving dichlorine monoxide in water; under standard aqueous conditions, anhydrous hypochlorous acid is currently impossible to prepare due to the readily reversible equilibrium between it and its anhydride.

2 HClO ⇌ Cl2O + H2O      K (at 0 °C) = 3.55×10−3 dm3 mol−1

The presence of light or transition metal oxides of copper, nickel, or cobalt accelerates the exothermic decomposition into hydrochloric acid and oxygen

2 Cl2 + 2 H2O → 4 HCl + O2

To learn more about exothermic decomposition click on the link below:

brainly.com/question/20089404

#SPJ4

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2.92 A 50.0-g silver object and a 50.0-g gold object are both added
Trava [24]

Answer:

82.9 mL  

Explanation:

1. Volume of silver

\begin{array}{rcl}\text{Density}&=& \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho&=& \dfrac{m}{V}\\\\V &=& \dfrac{m}{\rho}\\\\& = & \dfrac{\text{50.0 g}}{\text{10.49 g$\cdot$mL}^{-1}}\\\\& = & \text{4.766 mL}\\\end{array}\\\text{The volume of the silver is $\large \boxed{\textbf{4.766 mL}}$}

2. Volume of gold

\begin{array}{rcl}V& = & \dfrac{\text{50.0 g}}{\text{19.30 g$\cdot$mL}^{-1}}\\\\& = & \text{2.591 mL}\\\end{array}\\\text{The volume of the gold is $\large \boxed{\textbf{2.591 mL}}$}

3. Total volume of silver and gold

V = 4.766 mL + 2.591 mL = 7.36 mL

4 New reading of water level

V = 75.5 mL + 7.36 mL = 82.9 mL

3 0
4 years ago
What is the mole fraction of ethanol (C2H5OH) in a solution of 47.5 g of ethanol in 850 g of water?
Sveta_85 [38]

Answer:

0.021

Explanation:

that's the answer because I just did an exam that included this question and then the miss explained to me why this was the answer, but don't worry am 100% sure the answer is 0.021 .

6 0
3 years ago
5. Looking at the following 5 substances, Rank them in order of increasing solubility (the ability to dissolve in water). Briefl
dalvyx [7]

Answer

:just switch the first ones around

Explanation:

i d0nt know how to explain sorry

5 0
3 years ago
How many oxygen molecules in 2.3x10-⁸g of molecular oxygen ​
cluponka [151]

Answer:

6.321 × 10^22

Explanation:

Mass of Oxygen =

3.36

g

Molar mass of oxygen (

O

2

) = 16 x 2 =

32

g

m

o

l

−

Total molecules in oxygen = Mass in grams/Molar mass x

N

A

=

3.36

32

x

6.02

x

10

23

=

6.321

x

10

22

Note:

N

A

(Avagadro's number) =

6.02

x

10

23

Hope it helps...

3 0
3 years ago
You have 2000 g of radioactive substance with a half-life of 100 years four half-lives go by how many grams do you have the radi
Phoenix [80]

Answer:

after 4 half-lives, 125g is left

Explanation:

The half-life of a radioactive substance is the time it takes for the substance to decay by half its original mass.

In this example, we are asked to find the remaining mass after four half-lives.

What we will simply do is to reduce the starting mass by half, after each half-life decay. this is done as follows:

1st half-life decay: starting mass 2000g

final mass = 2000g ÷ 2 = 1000g

2nd half-life decay: starting mass = 1000g

final mass = 1000 ÷ 2 = 500

3rd half-life decay: starting mass = 500

final mass = 500 ÷ 2 = 250g

4th half-life decay: starting mass = 250g

final mass = 250 ÷ 2 = 125g

∴ after 4 half lives, 125g is left

3 0
3 years ago
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