Answer:
Electronegativity increases from left to right within a row of the periodic table
Explanation:
The density of an object ρ = 24 g/ml
<h3>Further explanation</h3>
Given
mass of an object = 120 g
volume = 5 ml
Required
The density
Solution
Density is a quantity derived from the mass and volume
Density is the ratio of mass per unit volume
Density formula:
ρ = density
m = mass
v = volume
Input the value :
ρ = 120 g : 5 ml
ρ = 24 g/ml
Answer:
(i) ΔU = 116 J
(ii) ΔU = 289 J
(iii) ΔU = 1 KJ
(iv) ΔU = 0 J
(v) ΔU = 3.25 KJ
Explanation:
first law:
(i) W = 153 J; Q = - 37 J ( Q ( - ), losing friction )
⇒ ΔU = 153 - 37 = 116 J
(ii) W = 289 J; Q = 0 ( insulated)
⇒ ΔU = W = 289 J
(iii) Q = 1 KJ , W = 0 ( isovolumetric process)
⇒ ΔU = Q = 1 KJ
(iv) isothermal ( constant temperature )
∴ ΔT = 0° ( isothermal )
⇒ ΔU = 0 J
(v) isobaric ( constant pressure )
⇒ ΔU = Q + W
∴ Q = 15.6 KJ
∴ W = - ∫ P dV = - P ΔV; W (-) the system performs a job and the volume increases
.
∴ P = 950 KPa * ( 1000 Pa / KPa ) = 950000 Pa = 950000 J/m³
∴ ΔV = 18 - 5 = 13 L * ( m³ / 1000 L ) = 0.013 m³
⇒ W = - ( 950000 J/m³) * ( 0.013 m³ ) = - 12350 J ( - 12.35 KJ )
⇒ ΔU = 15.6 KJ + ( - 12.35 KJ )
⇒ ΔU = 3.25 KJ
K2Cr2O7 + 14HCl → 2CrCl3 + 2KCl + 3Cl2 + 7H2O
the correct option is :
K2Cr2O7, because the oxidation number of Cr changes from +6 to +3.
<u>Oxidation number of Cr in K2Cr2O7 is:</u>
K2Cr2O7 = 2K + 2 Cr + 7 O
= 2(+1) + 2Cr + 7(-2)
= 2 + 2Cr -14
[total charge on K2Cr2O7 = 0], Hence;
2 + 2Cr -14 = 0
2Cr -12 = 0
2Cr = 12
Cr = 12/2
<u>Cr = +6</u>
<u>Oxidation number of Cr in CrCl3 is:</u>
CrCl3 = Cr + 3Cl = 0
Cr + 3(-1) = 0
Cr -3 = 0
<u>Cr = +3</u>
Hence Cr is changing its oxidation number from
+6 in K2Cr2O7 to +3 in CrCl3.
Since the oxidation number of Cr [ +6 → +3] is decreasing here,
Cr is getting reduced.
so K2Cr2O7 is an oxidizing agent,as it is getting itself reduced and oxidizes others.
