LAw of conservation of Energy is an important concept to solve this problem. The energy released is equal to the energy absorbed.
Ice undergoes latent heat. MEaning there is a change in phase but not temperature and the energy is solved by (enthalpy of fusion)(mass) = 333.5J/g)(8.5g). =2834.75J.
This is equal to the energy released by the water. The energy is computed by (mass)(specific heat of water)(temperature change) = (255g)(4.16J/gK)(T)
Final equation is:
2834.75 = 255(4.16)(T)
T = 2.67K
The molarity of a solution if it tale 12.0 grams of Ca(No3)2 is calculated as below
molarity = moles/volume in liters
moles = mass/molar mass = 12.0 g/ 164 g/mol = 0.073 moles
molarity is therefore = 0.073/0.105 = 0.7 M
Answer:
801 g
Explanation:
From the question given above, the following data were obtained:
Number of mole of Ba₃(PO₄)₂ = 1.33 moles
Mass of Ba₃(PO₄)₂ =?
Next, we shall determine the molar mass of Ba₃(PO₄)₂. This can be obtained as follow:
Molar mass of Ba₃(PO₄)₂ = (137.3×3) + 2[31 + (4×16)]
= 411.9 + 2[31 + 64]
= 411.9 + 2[95]
= 411.9 + 190
Molar mass of Ba₃(PO₄)₂ = 601.9 g/mol
Finally, we shall determine the mass of Ba₃(PO₄)₂. This can be obtained as follow:
Number of mole of Ba₃(PO₄)₂ = 1.33 moles
Molar mass of Ba₃(PO₄)₂ = 601.9 g/mol
Mass of Ba₃(PO₄)₂ =?
Mole = mass /Molar mass
1.33 = Mass of Ba₃(PO₄)₂ / 601.9
Cross multiply
Mass of Ba₃(PO₄)₂ = 1.33 × 601.9
Mass of Ba₃(PO₄)₂ = 801 g
Answer:
Freezing water can break rock without any change in the minerals that form the rock
Explanation:
<h2><u>Hope this helps→←</u></h2>
The question is incomplete. The complete question is stated below:
Two point charges are held at the corners of a rectangle as shown in the figure. The lengths of sides of the rectangle are 0.050 m and 0.150 m. Assume that the electric potential is defined to be zero at infinity.
a. Determine the electric potential at corner A.
b. What is the electric potential energy of a +3 µC charge placed at corner A?
Answer / Explanation:
a )V(A) = 1 / 4πe° ( - 5 5x10∧6C / 0.150m + 2x10∧6C / 0.050m )
The answer to the equation above is : = +6.0x10∧4 j/c
b) U(A) = qV(A)= (3.0x10∧6C) (6.0x10∧4 . j/c) =
The answer to the equation above is : =0.18 J
Explanation:
Where V(A) is equivalent to the electric potential
U(A) is equivalent to the electric potential energy
