Here is a comic strip created to explain the transformation of a substance as it changes phases.
Title: The Adventures of Mr. Ice Cube
Introduction: Mr. Ice Cube is a solid block of water. He lives in a freezer, where it is very cold.
Source of Thermal Energy: One day, Mr. Ice Cube is taken out of the freezer and placed in a hot cup of coffee. The hot coffee transfers thermal energy to Mr. Ice Cube, causing him to melt.
Particle Transformation from Solid to Liquid Phase: As Mr. Ice Cube melts, the particles in his solid structure start to move faster. They move so fast that they break free from the solid structure and become liquid particles.
Particle Transformation from Liquid to Gas Phase: As Mr. Ice Cube continues to heat up, the liquid particles start to move even faster. They move so fast that they escape from the liquid state and become gas particles.
Conclusion: Mr. Ice Cube has now transformed from a solid to a liquid to a gas. He is now a cloud of water vapor, floating in the air above the hot cup of coffee.
Find out more on thermal energy here: brainly.com/question/7541718
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Answer:I can't art but I envision a comic of a swimming pool with chlorine water in it. The bottom of the pool is black. The chlorine is happy and excited to protect the people going to swim in it. Then the sun comes out, warms the black tile, the water is heated and the chlorine is boiled into gas form. Unable to control its movement through the atmosphere, the large amount of chlorine from the in-ground pool infiltrates the home of its lovely owners, and they die from chlorine gas inhalation, as well as half the neighborhood.
The end.
(b) Calculate the mass of ammonia produced when 35.0g of nitrogen reacts with hydrogen.
Answer:
a) N2 (g) + H2 = 2 NH3
b) You have to state the mass of hydrogen
The partial pressure of Helium gas can be calculated by applying Daltons law of partial pressure and that comes out to be 227 mmHg.
Dalton's law of partial pressure helps in calculating the total pressure exerted by all the gases that are present in a mixture that is equal to the sum of partial pressure of all individual gases.
Partial pressure is the pressure that is exerted by one gas in a mixture of gases on the walls of the container. This law is applicable to only mixture of gases that are ideal gases. Ideal gases are the one which do not have any forces of attraction or repulsion between the particles of thee gas.
In our question
substituting all given values
=227mmHg
Thus the partial pressure of Helium gas is 227mmHg
To learn more about Dalton's law of partial pressure, here:
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Answer:
227 mmHg
Explanation:
Ptotal= PCO2 +PAr+PO2+PHe
PHe= Ptotal-PCO2 -PAr-PO2
PHe= (770 mmHg)-(128 mmHg)-(232 mmHg)-(183 mmHg)
PHe=227 mmHg
Answer:
1 litre of 1.0 M NaCl
Explanation:
When an ionic compound dissolves in water, it dissociates into ions. Consider the dissolution of sodium chloride in water;
NaCl(s) ------> Na^+(aq) + Cl^-(aq)
Hence, two solute particles are obtained from each formula unit of NaCl, a greater concentration of NaCl will contain a greater number of sodium an chloride ion particles.
Glucose is a molecular substance and does not dissociate in solution hence it yields a lesser number of particles in solution even at the same concentration as NaCl
The solution with the greatest number of solute particles is 1 litre of 1.0 M NaCl, as ionic compounds dissociate into individual ions, thus providing more particles per litre.
Given the details of the question, the solution that would be expected to contain the greatest number of solute particles would be 1 litre of 1.0 M NaCl. This is because when ionic compounds like sodium chloride are placed in water, they dissociate into individual ions. In the case of NaCl, it splits into two ions, sodium (Na+) and chloride (Cl-). Thus, a 1.0 M solution of NaCl would actually contain 2.0 moles of particles per litre because each formula unit of NaCl gives two particles. Covalently bonded molecules like glucose do not dissociate in solution, therefore, a 1.0 M glucose solution would have 1.0 mole of particles per litre.
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B. decrease down a group and increase across a period.
C. increase down a group and increase across a period.
D. increase down a group and remain constant across a period.
E. increase down a group and decrease across a period.
Answer: The correct answer is Option E.
Explanation:
Atomic radius is defined as the total distance measured from the nucleus of an atom to the outermost shell.
Trend down the group:
Moving from top to bottom, a new shell gets add up around the nucleus and the outermost shell gets far away from the nucleus. Due to this, the distance between the nucleus and outermost shell increases, which results in the increase of atomic radii of the atom.
Trend across the period:
Moving from left to right in a period, more and more electrons gets add up in the same shell. The attraction between the last electron and the nucleus increases. This results in the shrinkage of size of an atom. Thus, decreasing the atomic radii of the atom.
Hence, the correct answer is Option E.
E. increase down a group and decrease across a period. Atomic radii generally increase down a group due to extra electron shells and decrease across a period due to greater nuclear charge.
In general, the correct answer to this question is E: atomic radii increase down a group and decrease across a period on the Periodic Table. The atomic radii increase down a group due to the addition of extra electron shells. Each additional shell means a greater distance between the nucleus and the outermost electrons, which results in a larger atomic radius. On the other hand, as you move across a period from left to right, atomic radii typically decrease. This is due to an increase in positive charge in the nucleus which pulls the electrons closer, thus decreasing the atomic radius.
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Answer:
water and salt
Explanation:
In the FLVS salt will be your answer but technaically when a acid reacts with a base it can create salt and water
Answer:
Kc = 2.145 × 10⁻⁸¹
Explanation:
Let's consider the following reaction:
O₂(g) ⇄ 2O(g)
The standard Gibbs free energy for the reaction (ΔG°) can be calculated using the following expression:
ΔG° = Σnp. ΔG°f(p) - Σnp. ΔG°f(p)
where,
ni are the moles of products and reactants
ΔG°f(p) are the standard Gibbs free energy of formation of products and reactants
In this case,
ΔG° = 2 × ΔG°f(O) - 1 × ΔG°f(O₂)
ΔG° = 2 × 230.1 kJ/mol - 1 × 0 kJ/mol
ΔG° = 460.2 kJ/mol
With this information, we can calculate the equilibrium constant (Kc) using the following expression: