The mass of a sample of alcohol is found to be = m = 367 g
Hence, it is found out that by raising the temperature of the given product, the mass of alcohol would be 367 g.
Explanation:
The Energy of the sample given is q = 4780
We are required to find the mass of alcohol m = ?
Given that,
The specific heat given is represented by = c = 2.4 J/gC
The temperature given is ΔT = 5.43° C
The mass of sample of alcohol can be found as follows,
The formula is c = 
We can drive value of m bu shifting m on the left hand side,
m = 
mass of alcohol (m) = 
m = 367 g
Therefore, The mass of the given sample of alcohol is
m = 367g
It requires 4780 J of heat to raise the temperature by 5.43 C in the process which yields a mass of 367 g of alcohol.
The concentration of the sodium hydroxide solution in mol/l is 0.176 M.
Concentration is the abundance of a constituent divided by way of the overall volume of an aggregate. several sorts of mathematical descriptions may be outstanding: mass concentration, molar concentration, variety concentration, and extent awareness.
Given
V =25 ml = 0.025 L
M = 0.1
C₁ = 0.1
V₁ = 21.50 = 0.022 L
C₂ = ?
V₂ = 25 ml = 0.025 L
C₁V₁ = C₂V₂
C₂ = C₁V₁ / V₂
= 0.1 * 0.022 * 2 / 0.025
= 0.176 M
The concentration of a substance is the quantity of solute found in a given amount of solution. Concentrations are normally expressed in terms of molarity, defined because of the variety of moles of solute in 1 L of answer.
The Concentration of an answer is a measure of the quantity of solute that has been dissolved in a given amount of solvent or answer. A concentrated answer is one that has a rather huge quantity of dissolved solute.
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Answer:
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Explanation:
Mass of ethylene glycol = m = 100 g
Specific heat capacity of ethylene glycol = c = 3.5 J/g°C
Change in temperature of ethylene glycol = ΔT
Heat loss by the ethylene glycol = Q = 350 J


ΔT = 1°C
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
<span>The best answer is B. ICl experiences induced dipole-induced dipole interactions. Both iodine and chlorine belongs to the same group of the periodic table. Electronegativity decreases as you go down a group therefore Cl will have a greater attraction with the bond it forms with another atom. Dipole-dipole interactions form between I and Cl. For the Br2 molecule, no dipole occurs because they are two identical atoms. Therefore we will be expecting ICl will have a higher boiling point due to higher binding energy it forms.</span>
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>