Answer:
There is no exact answer for this question tbh.
Answer:
Yes, it is possible.
Explanation:
A diprotic acid is an acid that can release two protons. That's why it is called diprotic.
Monoprotic → Release one proton, for example Formic acid HCOOH
Triprotic → Releases three protons, for example H₃PO₄
Polyprotic → Release many protons, for example EDTA
it is a weak acid.
In the first equilibrum, it release proton, and the second is released in the second equilibrium. So the first equilibrium will have a Ka1
H₂A + H₂O ⇄ H₃O⁺ + HA⁻ Ka₁
HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂
The HA⁻ will work as an amphoterous because, it can be a base or an acid, according to this:
HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂
HA⁻ + H₂O ⇄ OH⁻ + H₂A Kb₂
Answer:
c. add coefficients as needed
Explanation:
A chemical equation is defined as the equation that shows changes in a chemical reaction. A chemical equation consist of reactant and product, reactant is at left side of the arrow and product is at right side of the arrow.
Reactant => Product
While balancing a chemical equation, the basic rule is to balance the coefficient as required. Coefficient represents the number of molecules and is used at front of a chemical symbol. Change in coefficient helps balance the number of atoms or molecules of the substances on both the sides of the arrow.
Subscripts are never allowed to change because it can change the chemical involved in the reaction.
Hence, the correct answer is "c. add coefficients as needed".
First, we have to get:
1- The heat required to increase T of ice from -50 to 0 °C:
according to q formula:
q1 = m*C*ΔT
when m is the mass of ice = mol * molar mass
= 1 mol * 18 mol/g
= 18 g
and C is the specific heat capacity of ice = 2.09 J/g-K
and ΔT change in temperature = 0- (-50) = 50°C
by substitution:
∴q1 = 18 g * 2.09 J/g-K *50°C
= 1881 J = 1.881 KJ
2- the heat required to melt this mass of ice is :
q2 = n*ΔHfus
when n is the number of moles of ice = 1 mol
and ΔHfus = 6.01 KJ/mol
by substitution:
q2 = 1 mol * 6.01 KJ/mol
= 6.01 KJ
3- the heat required to increase the water temperature from 0°C to 60 °C is:
q3 = m*C*ΔT
when m is the mass of water = 18 g
C is the specific heat capacity of water = 4.18 J/g-K
ΔT is the change of Temperature of water = 60°C - 0°C = 60°C
by substitution:
∴q3 = 18 g * 4.18 J/g-K * 60°C
= 4514 J = 4.514 KJ
∴the total change of enthalpy = q1+q2+q3
= 1.881 KJ +6.01 KJ + 4.514 KJ
= 12.405 KJ
Answer:
A. The neutrons and electrons are in the wrong place.
Explanation:
The atom's nucleus contains both protons and neutrons, whilst the electrons are arranged in shells around the nucleus.
