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malfutka [58]
3 years ago
15

Name two metals which will not liberate hydrogen fromdilute tetraoxosulphate(vi) acid.​

Chemistry
1 answer:
Cerrena [4.2K]3 years ago
6 0
Copper (Cu) and silver (Ag) would not liberate hydrogen because they are less reactive than hydrogen since their standard reduction potential is higher.
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Question 22 of 30
Black_prince [1.1K]

Answer:

C6H12O6+6O2--->6CO2+6H2O

Explanation:

So I went through all the answers and could not find the right one amongst. If I'm not wrong the reaction above is the reaction for respiration. The nearest answer is D but unfortunately the first reactant isn't in accordance with that which the question has given.

8 0
3 years ago
An electron can be added to halogen atom to force a halide ion with
Lilit [14]

An electron can be added to halogen atom to force a halide ion with 8 valence electrons

<h3>What is an atom?</h3>

An atom can be defined as the smallest part of an element which can take part in a chemical reaction.

However whenever, an electron is added to halogen atom to force a halide ion with 8 different valence electrons

So therefore; an electron can be added to halogen atom to force a halide ion with 8 valence electrons

Learn more about halogens:

brainly.com/question/18276987

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4 0
2 years ago
Which product of petroleum is normally used as fuel in cars and light vehicles​
Elena L [17]

Answer:

mostly gasoline in cars nowadays

8 0
3 years ago
Read 2 more answers
A covalent bond forms a(n) ______________.
DiKsa [7]
The answer is Molecule. After a pair of shared electrons orbit around the nuclei of both atoms
3 0
2 years ago
Read 2 more answers
determine the frequency and wavelength (in nm) of the light emitted when the e- fell from n=4 and n=2
Lostsunrise [7]

Answer:

Frequency = 6.16 ×10¹⁴ Hz

λ = 4.87×10² nm

Explanation:

In case of hydrogen atom energy associated with nth state is,

En =  -13.6/n²

For n = 2

E₂ = -13.6 / 2²

E₂ = -13.6/4

E₂ = -3.4 ev

Kinetic energy of electron = -E₂ = 3.4 ev

For n = 4

E₄ = -13.6 / 4²

E₄ = -13.6/16

E₄ = -0.85 ev

Kinetic energy of electron = -E₄ = 0.85 ev

Wavelength of radiation emitted:

E = hc/λ = E₄ - E₂

hc/λ = E₄ - E₂

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev   - (-3.4ev )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J

λ = 4.87×10⁻⁷ m

m to nm:

4.87×10⁻⁷ m ×10⁹nm/1 m

4.87×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /4.87×10⁻⁷ m

Frequency = 6.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 6.16 ×10¹⁴ Hz

3 0
3 years ago
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