The substance that releases the greatest amount of ions will have the greatest attractive forces within its solution, resulting in a reduced freezing point.
K₂SO₄ yields 3 ions
NH₄I yields 2 ions
CoCl₃ yields 4 ions
Freezing points:
CoCl₃ < K₂SO₄ < NH₄I
Answer:
Firstly, Let's experiment this !
Experiment 1 :
159.446g - 124.966g = 34.48g
34.48g = The mass of Mineral oil.
The density of the mineral oil = M/V = 34.48g/40mL = 0.862g/cm³.
Experiment 2 :
124.966 + 18.173 = 143.139 = The mass of solid + cylinder.
124.966 + 50.952 = 175.918 = The mass of solid + cylinder + Mineral water.
175.918 - 143.139 = 32.779 = The mass of added mineral oil.
Explanation:
Now we have to find the volume of the added mineral oil using the density from experiment 1.
V = 32.779g/0.862g/cm³ = 38.02668213mL
Since we found the volume of the solid, we then have to subtract the added mineral oil volume from the total volume from experiment 1.
Volume of solid = 40-38.02668213 = 1.97331787mL
Density of solid = 18.713g/1.97331787mL = 9.483013499g/cm^3
1.97331787 = (4/3)(3.14)r³
1.97331787*(3/4)(3.14) = .4713338861
.4713338861 = r
³
r = 0.7782328425158433
r = 0.78
Now that's our final answer ! r = 0.78
1. 2Al(s)+6HCl(aq)⇒2AlCl₃(aq)+3H₂(g)
2. 2AgNO₃ (aq) + Cu (s)⇒Cu(NO₃)₂ (aq) + 2Ag (s)
3. 2C₃H₈O(l) + 9O₂(g) ⇒ 6CO₂(g) + 8H₂O(g)
<h3>Further explanation</h3>
There are several reactions that can occur in a chemical reaction: single replacement, double replacement, synthesis, decomposition or combustion, etc.
1.Al(s)+HCl(aq)⇒AlCl₃(aq)+H₂(g)
type : single replacement
balance :
2Al(s)+6HCl(aq)⇒2AlCl₃(aq)+3H₂(g)
2. AgNO₃ (aq) + Cu (s) ⇒ Cu(NO₃)₂ (aq) + Ag (s)
type : single replacement
balance :
2AgNO₃ (aq) + Cu (s)⇒Cu(NO₃)₂ (aq) + 2Ag (s)
3. C₃H₈O + O₂ ⇒ CO₂ + H₂O
type : combustion of alcohol
balance :
2C₃H₈O(l) + 9O₂(g) ⇒ 6CO₂(g) + 8H₂O(g)
Answer:
The speed of the 60.0 kg skater should be 0.281 m/s
Explanation:
<u>Step 1: </u>Data given
Mass of skater 1 = 45.0 kg
speed of skater 1 = 0.375 m/s
Mass of skater 2 = 60.0 kg
<u>Step 2:</u> Calculate the speed of skater 2
To solve this problem, we will use 'Conservation of momenton'. This means the momentum before the push equals the momentum after.
momentum p = m*v
Momentum p(before) = momentum p(after)
m1*v1 = m2 * v2
⇒ with m1 = mass of skater 1 = 45.0 kg
⇒ with v1 = the velocity of skater 1 = 0.375 m/s
⇒ with m2 = the mass of skater 2 = 60.0 kg
⇒ with v2 = the velocity of skater 2 = TO BE DETERMINED
45.0 * 0.375 = 60.0 * v2
v2 = (45.0*0.375)/60
v2 = 0.281 m/s
The speed of the 60.0 kg skater should be 0.281 m/s
