Explanation:
Mass of solute = 10.0 g
mass of solvent(water) = m
Volume of solvent( water) = v = 100.0 mL
Density of water= d = 
Mass of solution(M) = Mass of solute + mass of solvent
M = 10.0 g + 100.0 g = 110.0 g
Volume of the solution = V = 113 mL
Density of the solution = D
The density of the solution is 0.9734 g/ml.
Moles of phosphoric acid = 
Moles of water = 
Mole fraction of phosphoric acid =
Mole fraction of water =
![[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}](https://tex.z-dn.net/?f=%5BMolarity%5D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%28L%29%7D%7D)
Moles of phosphoric acid = 0.1020 mol
Volume of the solution = V = 113 mL = 0.113 L ( 1 mL = 0.001 L)
Molarity of the solution :
![[Molality]=\frac{\text{Moles of solute}}{\text{Mass of solvent(kg)}}](https://tex.z-dn.net/?f=%5BMolality%5D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BMass%20of%20solvent%28kg%29%7D%7D)
Moles of phosphoric acid = 0.1020 mol
Mass of solvent(water) = m =100.0 g = 0.100 kg ( 1 g = 0.001 kg)
Molality of the solution :
Answer:
<h2>Transverse wave,</h2>
Explanation:
<h3>motion in which all points on a wave oscillate along paths at right angles to the direction of the wave's advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves.</h3>
Answer:
Sodium chloride solution:
First you need to calculate the mass of salt needed (done in the explanation), which is 58.44g. Then it have to be weighted in an analytical balance in a weighting boat and then transferred into a 2L volumetric flask that is going to be filled until the mark with distilled water.
Sulfuric acid dilution:
First you need to calculate the volume needed (done in the explanation), it is 16.6 mL. Using a graduated pipette one measures this volume and transfer it into a 2L volumetric flask that is already half filled with distilled water, and then one fills it until its mark.
Explanation:
Sodium chloride solution:
Each liter of a 0.500M solution has half mol, so 2L of said solution has 1 mol of salt. Sodium chloride molar mass is 58.44g/mol, so in 2L of solution there is 58.44g of salt. That`s the mass that`s going to be weighted and transferred to a 2L volumetric flask.
Sulfuric acid dilution:
This is the equation for dilution of solutions:
Where "c1" stands for the initial concentration (stock solution concentration), "v1" for the initial volume (volume of stock solution used), "c2" for the desired concentration and "v2" for the desired volume.
When we are diluting from a stock solution we want to know how much do we have to pipette from the stock solution into our volumetric flask. We do so by isolating the "v1" term from the dilution equation:
in this case that would be:
Answer:
Q = 4019.4 J
Explanation:
Given data:
Mass of ice = 20.0 g
Initial temperature = -10°C
Final temperature = 89.0°C
Amount of heat required = ?
Solution:
specific heat capacity of ice is 2.03 J/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 89.0°C - (-10°C)
ΔT = 99°C
Q = 20.0 g ×2.03 J/g.°C × 99°C
Q = 4019.4 J
