Answer:
The temperature is 30,92K
Explanation:
We use the formula PV=nRT. We convert the unit of pressure in kPa into atm.
101,325kPa----1atm
121kPa-------x=(121,3kPax 1 atm)/101,325kPa=1, 2 atm
PV=nRT---->T= (PV)/(RT)
T=(1,2 atm x 3L)/(1,42 mol x 0,082 l atm/K mol )= 30, 91721058 K
The 2 L of sucrose stock solution would contain similar
concentration with the 100 mL aliquot. Therefore the concentration of aliquot
is still 2 M.
The molar mass of sucrose is 342.3 g / mol. Therefore the
mass in a 100 mL (0.1 L) aliquot is:
mass = (2 mol / L) * 0.1 L * (342.3 g / mol)
<span>mass = 68.46 g</span>
Answer:
1·199 J
Explanation:
Given
Mass of water = 0·814 g = 0·814 × 
kg
Increase in temperature = 0·351 °C
Let the amount of heat added be Q J
Formula for heat added is
<h3>Q = m × s × ΔT</h3>
where Q is the amount of heat transferred
m is the mass
s is the heat capacity
ΔT is the change in temperature
Heat capacity of water = 4200 J/kg °C
Applying the formula for heat added
Q = 0·814 × × 4200 × 0·351 = 1·199 J
∴ Amount of heat added = 1·199 J
In NaMnO₄, Mn has the highest oxidation number.
The question is incomplete, the complete question is;
Which of the following species contains manganese with the highest oxidation number?
A) Mn
B) MnF₂
C) Mn₃(PO₄)₂
D) MnCl₄
E) NaMnO₄
In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.
1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.
2) For MnF₂;
Mn has an oxidation number of +2
3) For Mn₃(PO₄)₂
Mn has an oxidation number of +2
4) For MnCl₄
Mn has an oxidation number of +4
5) For NaMnO₄
Mn has an oxidation number of +7
Hence in NaMnO₄, Mn has the highest oxidation number.
Learn more: brainly.com/question/10079361
