Answer:
A cold front is a boundary or zone that separates two air masses. The cooler more dense mass replaces the warmer.
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Explanation:
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.
Answer:
1. The α particles were repelled by electrons.
Explanation:
The gold foil experiment was performed by Rutherford and his research group in 1911 (at the beginning of the 20th century). In this experiment, α particles were bombed to gold foils, and films were placed surround it to collect the particles.
It was observed that most of the particles passed through of the foil undeflected, and for that, Rutherford stated that the atom was a "huge empty". Some particles were deflected, because they're attracted to the electrons at the electrosphere, and a small number of particles were complete deflected to the origin because they chocked with the small positive nuclei.
Thus, the experiment suggested the nuclear model of the atom, called the planetary model, that was improved after by Bohr and other scientists in the quantum model.
Answer:
So there is83.6g CO2 produced
Explanation:
Burning carbon with air has the following equation
C + O2 → CO2
For 1 mol Carbon, we have 1 mol O2 and 1 mol CO2
Step 2: Calculating moles
mole C = 22.8g / 12g/mole
Mole C = 1.9 mole
1.9 mole C will completely react
Since for each mole C there is 1 mole O2 and 1 mole CO2
This means there will also react 1.9 mole of 02, to be formed 1.9 mole of CO2
mole CO2 = mass CO2 / Molar mass CO2
mass CO2 = 1.9 mole CO2 * 44g/mole =<u>83.6g CO2</u>
In this reaction 18.2 g of O2 remained unreacted
we can control this: 79g - 18.2 g = 60.8g
1.9 mole * 32g/mol = 60.8g
So there is83.6g CO2 produced
Answer:
Empirical formula of C₈H₈ = CH
Explanation:
Data Given:
Molecular Formula = C₈H₈
Empirical Formula = ?
Solution
Empirical Formula:
Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.
So,
tha ration of the molecular formula should be divided by whole number to get the simplest ratio of molecule
C₈H₈ Consist of Carbon (C), and Hydrogen (H)
Now
Look at the ratio of these two atoms in the compound
C : H
8 : 8
Divide the ratio by two to get simplest ratio
C : H
8/8 : 8/8
1 : 1
So for the empirical formula is the simplest ratio of carbon to hydrogen 1 : 1
So the empirical formula will be
Empirical formula of C₈H₈ = CH
