Answer:
150.1 mL
Explanation:
Step 1: Given data
- Density of benzene (ρ): 0.879 g/mL
- Mass of the sample of benzene (m): 131.9 g
- Volume of the sample of benzene (V): ?
Step 2: Calculate the volume of the sample of benzene
Density is an intrinsic property. It is equal to the quotient between the mass and the volume of the sample of benzene.
ρ = m/V
V = m/ρ
V = 131.9 g/(0.879 g/mL)
V = 150.1 mL
Answer:
Glucose
Explanation:
Glucose is a source of energy and it is stored in the body (in the form of glycogen) and it can also dissolve in water.
A=Mass number=24
N=neutrons=13
Z=atomic number.
A=Z+N
24=Z+13
Z=24-13
Z=11
The atomic number is 11, and this atom is sodium.
setup 1 : to the right
setup 2 : equilibrium
setup 3 : to the left
<h3>Further explanation</h3>
The reaction quotient (Q) : determine a reaction has reached equilibrium
For reaction :
aA+bB⇔cC+dD
![\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Comparing Q with K( the equilibrium constant) :
K is the product of ions in an equilibrium saturated state
Q is the product of the ion ions from the reacting substance
Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)
Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium
Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)
Keq = 6.16 x 10⁻³
Q for reaction N₂O₄(0) ⇒ 2NO₂(g)
![\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Setup 1 :

Q<K⇒The reaction moved to the right (products)
Setup 2 :

Q=K⇒the system at equilibrium
Setup 3 :

Q>K⇒The reaction moved to the left (reactants)
Answer:
%KCl = 7.05%
%Water = 92.95%
Explanation:
Step 1: Given data
- Mass of KCl (solute): 36 g
- Mass of water (solvent): 475 g
Step 2: Calculate the mass of the solution
The mass of the solution is equal to the sum of the masses of the solute and the solvent.
m = 36 g + 475 g = 511 g
Step 3: Calculate the mass percentage of the solution
We will use the following expression.
%Component = mComponent/mSolution × 100%
%KCl = 36 g/511 g × 100% = 7.05%
%Water = 475 g/511 g × 100% = 92.95%