Answer:
layer B is liquid because of the inner members and layer A is solid because of the outer members
Answer:
5.00 mol Mg
10.0 mol Cl
40.0 mol O
Explanation:
Step 1: Given data
Moles of Mg(ClO₄)₂: 5.00 mol
Step 2: Calculate the number of moles of Mg
The molar ratio of Mg(ClO₄)₂ to Mg is 1:1.
5.00 mol Mg(ClO₄)₂ × 1 mol Mg/1 mol Mg(ClO₄)₂ = 5.00 mol Mg
Step 3: Calculate the number of moles of Cl
The molar ratio of Mg(ClO₄)₂ to Cl is 1:2.
5.00 mol Mg(ClO₄)₂ × 2 mol Cl/1 mol Mg(ClO₄)₂ = 10.0 mol Cl
Step 4: Calculate the number of moles of O
The molar ratio of Mg(ClO₄)₂ to Cl is 1:8.
5.00 mol Mg(ClO₄)₂ × 8 mol O/1 mol Mg(ClO₄)₂ = 40.0 mol O
Explanation:
Scientist Evidence Model
Dalton Gases indivisible, solid and spheres
J.J Thomson Deflected beam Negative charges evenly scattered
through positively charged mass of
matter.
Rutherford Deflection of alpha atomic model
particles passing
through the gold foil
The emboldened words are the answer.
- John Dalton proposed the first model of the atoms by his works on gases. He postulated the Dalton's law of partial pressure.
- He suggested that gases are made of tiny particles called atoms.
- J.J Thomson proposed the plum pudding model of the atom in which the charges are evenly scattered through the positively charged mass of matter.
- The gold foil experiment by Rutherford brought the atomic model of the atoms in the front-light.
- The model suggests a small positively charged center which the mass of the atom.
- The outer space is occupied by the electrons.
Learn more:
Rutherford brainly.com/question/1859083
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Heat = 1.74 kJ
<h3>Further explanation</h3>
Given
melts at 328 ℃ + 273 = 601 K
mass = 23 g = 0.023 kg
initial temperature = 297 K
Final tmperature = 702 K
Required
Heat
Solution
1. raise the temperature(297 to 601 K)
c of lead = 0.130 kJ/kg K
Q = 0.023 x 0.13 x (601-297)
Q = 0.909 kJ
2. phase change(solid to liquid)
Q = m.Lf (melting/freezing)
Q = 0.023 x 23 kj/kg = 0.529 kJ
3. raise the temperature(601 to 702 K)
Q = 0.023 x 0.13 x (702-601)
Q = 0.302 kJ
Total heat = 1.74 kJ