Answer:
chemical change occur when a substance combines with another to form a new substance.
Explanation:
hope this helps ;)
Answer:
It is the solution of a mixture of a weak base and a salt of this weak base with a strong acid.
Explanation:
Quantum numbers are the conserved quantities in the system that expresses the position of the electron. The second quantum number describes the electron subshells.
<h3>What are quantum numbers?</h3>
Quantum numbers are the energy and the position defining values of the electron in the atomic structure. There are four basic quantum numbers, principal, magnetic, azimuthal, and spin quantum.
The second quantum number (I) is the angular quantum number that describes the magnitude of the subshells (s,p, d, and f) or the type of the subshells. It is also known as the Azimuthal quantum number.
Therefore, the second quantum number shows the electronic subshells.
Learn more about quantum numbers here:
brainly.com/question/4889088
#SPJ1
Answer:
CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
Explanation:
The reaction between solid calcium carbonate and a mineral acid such as aqueous HCl is a neutralization reaction and occurs with the evolution of CO2 gas.
The balanced equation is given below
CaCO3 + 2HCl ---> CaCl2 + H20 + CO2
The product CaCl2 is water soluble which accounts for why the stain is removed, while CO2 gas escapes away from the reaction surface.
Answer:
E°red for Tl³⁺(aq) | Tl⁺(aq) is 0.78 V
Explanation:
According to cell reaction we can predict which one cathode half cell and which one anode.
We know left hand side electrode where oxidation occur acts as an anode and right hand side electrode where reduction take place acts as an cathode.
Hence Cr²⁺ converted into Cr³⁺ since oxidation is take place where as Ti³⁺ converted into Ti⁺ means reduction take place.
<u>Anode half cell </u>is Cr³⁺(aq) | Cr²⁺(aq) is E⁰red = -0.41 V
<u>Cathode half cell</u> is Ti³⁺(aq) | Ti⁺(aq) is E⁰red = ?
E⁰cell = E⁰cathode - E⁰anode
⇒ E⁰cathode = E⁰cell + E⁰anode
⇒ E⁰cathode = 1.19 + (- 0.41) volt = 0.78 V
