Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
Answer:
Explanation:
mass fraction N₂ : He : CH₄ : C₂H₆ : : 15 : 5 : 60 : 20
mole fraction N₂ : He : CH₄ : C₂H₆ : : 15/28 : 5/4 : 60/16 : 20/30
mole fraction N₂ : He : CH₄ : C₂H₆ : : .5357 : 1.25 : 3.75 : .67
Total mole fractions = .5357 + 1.25 + 3.75 + 0.67 = 6.2057
mole fraction of N₂ = .5357 / 6.2057 = .0877
mole fraction of He = 1.25 / 6.2057 = .20
mole fraction of CH₄ = 3.75 / 6.2057 = .6043
mole fraction of C₂H₆ = .67 / 6.2057 = .108
Partial pressure = total pressure x mole fraction
Partial pressure of N₂ = 1200 kPa x .0877 = 105.24 kPa
Partial pressure of He = 1200 kPa x .20 = 240 kPa
Partial pressure of CH₄ = 1200 kPa x .6043 = 725.16 kPa
Partial pressure of C₂H₆ = 1200 kPa x .108 = 129.6 kPa
Answer:
Explanation:
= Mass of metal = 19 g
= Specific heat of the metal
= Temperature difference of the metal = 
V = Volume of water = 150 mL = 
= Density of water = 
= Specific heat of the water = 4.186 J/g°C
= Temperature difference of the water = 
Mass of water
Heat lost will be equal to the heat gained so we get
The specific heat of the metal is .
shared to form an ionic bond.
I hope this helps!
