The most condensed state of matter is A. Solid Matter
Answer:
The correct answers are "They travel slower than P waves", "They result in much ground motion" and " They are produced by P and S waves".
Explanation:
A surface wave can travel through the existing interface between Earth and air, as well as between Earth and water. A clear example of this type of waves are Love and Rayleigh waves. They travel slower than primary waves, produce large movements in the ground and are produced by primary and secondary waves.
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Answer:
A. 0.2395 w/w %
B. 2394ppm
Explanation:
A. To find concentrationin percent by mass of the solution we need to calculate mass of glycerol and mass of water. The formula is:
Mass glycerol / Total mass * 100
<em>Mass glycerol:</em>
The solution is 2.6x10⁻²moles / L. As there is 1L of solution there are 2.6x10⁻² moles of glycerol. In mass (Using molar mass glycerol: 92.09g/mol):
2.6x10⁻² moles of glycerol * (92.09g / mol) = 2.394g glycerol
<em>Mass of water:</em>
998.9mL and density = 0.9982g/mL:
998.9mL * (0.9982g/mL) = 997.1g of water.
That means percent by mass is:
% by mass: 2.394g / (997.1g + 2.394g) * 100 = 0.2395 w/w %
B. Parts per million are mg of glycerol per L of solution. As in 1L there are 2.394g. In mg:
2.394g * (1000mg / 1g) = 2394mg:
Parts per million: 2394mg / L = 2394ppm
Answer:
43.96secs
Explanation:
M1 = molar mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol
t1 = time for NH3 to diffuse = 30secs
M2 = molar mass of HCl = 1 + 35.5 = 36.5g/mol
t2 = time for HCl to diffuse =?
From Graham's law of diffusion:
t2/t1 = √(M2/M1)
t2/30 = √(36.5/17)
t2/30 = 1.465
Cross multiply
t2 = 30 x 1.465
t2 = 43.96secs
Therefore it will take 43.96secs for the damp blue litmus paper to change colour for HCl.
Answer:
1. 25 moles water.
2. 41.2 grams of sodium hydroxide.
3. 0.25 grams of sugar.
4. 340.6 grams of ammonia.
5. 4.5x10²³ molecules of sulfur dioxide.
Explanation:
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In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:
1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:
2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:
3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:
4. Here, since the molar mass of ammonia is 17.03 g/mol:
5. Here, since the molar mass of SO2 is 64.06 g/mol:
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