Answer:
The ball will fly tangential to the original circle
Explanation:
The image here is missing, however we can still answer to the question.
In fact, the circular motion of the ball when it is tied to the rope is a combination of two separate effects:
1- The centripetal force, in the form of the tension in the rop, that pulls the ball at any time towards the centre of the circular path
2- The inertia of the ball, which tends to continue its motion in a straight direction, tangential to the circle and perpendicular to the direction of the centripetal force
When child let the string go, there is no more tension in the string acting on the ball, and therefore, there is no longer a centripetal force.
As a result, number 1) disappears, and therefore there is only the inertia of the ball that will determine its motion: and therefore, the ball will continue its motion straight in a direction tangential to the original circle.
Answer:
(D) Na₂SO₄•10H₂O (M = 286).
Explanation:
- The depression in freezing point of water by adding a solute is determined using the relation:
<em>ΔTf = i.Kf.m,</em>
Where, ΔTf is the depression in freezing point of water.
i is van't Hoff factor.
Kf is the molal depression constant.
m is the molality of the solute.
- Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).
- van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.
(A) CuSO₄•5H₂O:
CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.
So, i = dissociated ions/no. of particles = 2/1 = 2.
B) NiSO₄•6H₂O:
NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.
So, i = dissociated ions/no. of particles = 2/1 = 2.
(C) MgSO₄•7H₂O:
MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.
So, i = dissociated ions/no. of particles = 2/1 = 2.
(D) Na₂SO₄•10H₂O:
Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.
So, i = dissociated ions/no. of particles = 3/1 = 3.
∴ The salt with the high (i) value is Na₂SO₄•10H₂O.
So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.
8.5L H2O (1 mol H2O/22.4L)=.37946 mol H2O
.3796 mol H2O(1 mol CH4/2 mol H2O)=.18973 mol CH4
.18973 mol CH4(22.4L/1mol CH4)=4.25L CH4 gas
Answer:
Chlorine gas is the limiting reactant
11.57 g
Explanation:
The equation of the reaction is;
2Al(s) + 3Cl2(g) -----> 2AlCl3(s)
The limiting reactant will produce the least number of moles of AlCl3.
For Al
Number of moles reacted = 2.43 g/27 g/mol = 0.09 moles
2 moles of Al yields 2 moles of AlCl3
0.09 moles of Al yields 0.09 * 2/2 = 0.09 moles of AlCl3
For Cl2
9.22g/71g/mol = 0.13 moles
3 moles of Cl2 yields 2 moles of AlCl3
0.13 moles of Cl2 yields 0.13 * 2/3 = 0.087 moles of AlCl3
Hence Cl2 is the limiting reactant
b) Mass of product produced = 0.087 moles of AlCl3 * 133g/mol = 11.57 g
