Jack's model includes one object consisting of two molecules and one object consisting of six molecules. He put 16 energy cubes in each object and stated that his model showed the two objects at equilibrium.
Explain what is wrong with Jack’s model.
The mistake is when you book them together the molecules have different energy and the system is not in equilibrium.
Given data:
Jack's model consists of 1 object = 2 molecules.
And, object 2 = 6 molecules.
And there were 16 energy cubes kept in each object.
As per the given problem, the model developed by Jack has the same energy for each of the two objects, but as each object is made up of a different number of molecules, in the system with more molecules, object 2, each one has approximately 2.4 less energy and the molecules of Object 1 have an E / 2 energy.
Thus, we can conclude that the mistake is when you book them together the molecules have different energy and the system is not in equilibrium.
Answer: b. Fiona is correct because the diagram shows two individual simple machines.
Explanation:
A mechanical device using which we can change the direction or magnitude of force applied is known as simple machine.
For example, in the given diagram there are two individual simple machines.
The machine helps in changing the direction or magnitude of force applied by the man. As a result, it becomes easy for him to carry different things easily from one place to another.
Thus, we can conclude that the statement Fiona is correct because the diagram shows two individual simple machines, is correct.
15 moles.Explanation:Hello,In this case, the undergoing chemical reaction is:Clearly, since carbon and oxygen are in a 1:1 molar ratio, 15 moles of carbon will completely react with 15 moles of oxygen, therefore 15 moles of oxygen remain as leftovers. In such a way, since carbon and carbon dioxide are also in a 1:1 molar ratio, the theoretical yield of carbon dioxide is 15 moles based on the stoichiometry:Best regards.
Answer: 7.8125 g
Explanation: Given:
Original amount (N₀) = 500 g
Number of half-lives (n) = 9612/1602 = 6
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^6 × 500
N = 0.015625 × 500
N = 7.8125 g
Answer: The product formed is potassium chloride.
Explanation:
Precipitation reaction is defined as the chemical reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.
The chemical equation for the reaction of potassium phosphate and magnesium chloride follows (look at the picture)
2 moles of aqueous solution of potassium phosphate reacts with 3 moles of aqueous solution of magnesium chloride to produce 1 mole of solid magnesium phosphate and 6 moles of aqueous solution of potassium chloride.
Answer: B. carbon tetrachloride, CCI4
Explanation: The other options are incorrect. Let's write the correct formulas:
A. Diarsenic pentoxide - As2O5
C. Sodium dichromate - Na2Cr2O7
D. magnesium phosphide - Mg3P2
Explanation: Acids and bases can be classified in terms of hydrogen ions or hydroxide ions, or in terms of electron pairs. (look at the picture)
Let us note that from the pH scale, a pH of;
0 - 6.9 is acidic
7 is neutral
8 - 14 is basic
a. [H+] = 6.0 x 10-10M
pH= 9.22 is basic
b. [OH-] = 30 × 10-2M
pH = 13.5 is basic
C. IH+1 = 20× 10-7M
pH = 0.56 is acidic
d. [OH-] = 1.0 x 10-7M
pH = 7 is neutral
Answer: 306 L
Explanation: Using the ideal gas law,
PV = nRT
where R = 0.08206 L•atm/(mol•°K), solving for n gives
n = PV/(RT)
n = (845 mmHg) (270 L) / ((0.08206 L•atm/(mol•°K)) (24 °C))
Convert the given temperature to °K and the given pressure to atm:
24 °C = (273.15 + 24) °K ≈ 297.2 °K
(845 mmHg) × (1/760 atm/mmHg) ≈ 1.11 atm
Then the balloon contains
n = (1.11 atm) (270 L) / ((0.08206 L•atm/(mol•°K)) (297.2 °K))
n ≈ 12.3 mol
of He.
Solve the same equation for V :
V = nRT/P
Convert the target temperature to °K:
-50 °C = (273.15 - 50) °K = 223.15 °K
Then the volume under the new set of conditions is
V = (12.3 mol) (0.08206 L•atm/(mol•°K)) (223.15 °K) / (0.735 atm)
V ≈ 306 L
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