Answer:
You can find in the given attachemnet
Answer:
Cells of a multicellular organism
Answer:
The radial velocity curve describes how fast a star is moving in its orbit around a center of mass ( m )
Curve amplitude : This is the maximum value of the radial velocity curve
Radial velocity shape ; The shape of Radial velocity curve is parabolic in nature
Orbital period : Orbital period is the time taken by the star to make one complete rotation in its orbit
Explanation:
The radial velocity curve describes how fast a star is moving in its orbit around a center of mass ( m ) while Curve amplitude is the maximum value of the radial velocity curve also The shape of Radial velocity curve is parabolic in nature. and Orbital period is the time taken by the star to make one complete rotation in its orbit
Answer:
The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.
Explanation:
For N2,
Pressure(P₁)=125 kPa
Volume(V₁)=15·1 L
Temperature (T₁)=25°C=25+273 K=298 K
Similarly, for Oxygen,
Pressure(P₂)= 125 kPa
Volume(V₂)= 44.3 L
Temperature(T₂)=25°C= 298 K
Then, for the mixture,
Volumeof the mixture( V)= 6.25 L
Pressure(P)=?
Temperature (T)= 51°C = 51+273 K=324 K
Then, By Combined gas laws,
or, 
or, 
or, 
∴P=1291.85 kPa
So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.
Answer:
202 g/mol
Explanation:
Let's consider the neutralization between a generic monoprotic acid and KOH.
HA + KOH → KA + H₂O
The moles of KOH that reacted are:
0.0164 L × 0.08133 mol/L = 1.33 × 10⁻³ mol
The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 × 10⁻³ moles.
1.33 × 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:
0.2688 g/1.33 × 10⁻³ mol = 202 g/mol
