D. Physical change
Cause you are the one doing the change to it and physical change means You.
Answer: It passes through both mantle and core, but are slowed and refracted at the mantle / core boundary at a depth of 2900 km.
Explanation:
Answer:
a, d and e. are true.
Explanation:
The reaction that occurs is:
Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2NaCl
In ideal conditions, the percent yield of the reaction must be 100%. All explanations about why the student could not collect all precipitate are right:
a. The combined reactants were not stirred before filtering the precipitate. Not stirring could not promote all the reaction. <em>TRUE.</em>
b. The student did not completely dry the precipitate before weighing it. If the student don't dry the precipitate, the mass of precipitate must be higher producing a percent yield > 100%. <em>FALSE.</em>
c. The precipitate was not washed prior to drying. Produce more mass. <em>FALSE.</em>
d. A rubber policeman was not used to scrape precipitate from the beaker. If the student doesn't collect all the precipitate the percent yield could be < 100%.. <em>TRUE.</em>
e. The filter paper was not wetted with water prior to filtering the precipitate. <em>TRUE. </em>If you don't wet the filter paper you can lose a part of precipitate from the walls of this one.
Answer:
A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.
Explanation:
A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.
1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J
ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J
B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂
= -4.3J -(-6.0 J)= 1.7 J. Since U₁ - U₂ > 0 , U₂ < U₁ , so the second system's internal energy increase less and is lower in energy and is more stable.
<span>According to Faraday's Law,
W = I t e / F
Where,
W = Amount deposited = ?
I = Current = 3 A
t = Time = 965 s
e = Chemical equivalence = 52/3 = 17.33
F = Faraday's Constant = 96500
Putting Values,
W = (3 × 965 × 17.33) ÷ 96500
W = (50170.35) ÷ 96500
W = 0.52 g
As we know,
Moles = mass ÷ M.mass
Putting values,
Moles = 0.52 ÷ 52
Moles = 0.01</span>