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stepan [7]
2 years ago
6

Select the Coral Reef Tab. Make sure that you have clicked Return to original settings and Restart. Set the Ocean Temperature to

33 Degrees Celsius. Click Advance year 10 times. What changes do you notice?
Chemistry
1 answer:
Alex17521 [72]2 years ago
6 0

Answer:

What changes do you notice? The white you see have undergone coral bleaching. At high temperatures, corals may lose their zooxanthellae, causing corals to lose their color and their main source of food. Once bleaching occurs, the coral colony usually dies.

Explanation:

study well

Arigato

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An endothermic reaction needs energy to proceed, such energy is usually taken from the environment surrounding the reaction. In the typical case this energy is expressed as  heat. Heat is an state of atomic activity, that energy is transferred to an ENDOthermic reaction so the initial threshold of reaction is overcome and the final reaction can occur.
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When 3.18g of copper(ii)oxide was carefully heated in a stream of dry hydrogen, 2.54g of copper and 0.72g of water was formed. D
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An engineering team has a goal of developing a bicycle frame that is both
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4 years ago
Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi
Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g)     ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

[]=\frac{P}{R.T}

<em>Calculate ΔG at 25°C given the following sets of partial pressures.</em>

<em>Part A  130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.</em>

[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M

[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

<em>Part B  5.0atm SO₂, 3.0atm O₂, 30atm SO₃  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M

[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M

[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

<em>Part C Each reactant and product at a partial pressure of 1.0 atm.  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0
3 years ago
In the explosion of a hydrogen-filled balloon, 0.80 g of hydrogen reacted with 6.4 g of oxygen. How many grams of water vapor ar
mylen [45]

 0.4 moles H2O x (18.0 g / mole) = 7.2 g H2O

Explanation:

write the values given in the question

In an explosion of hydrogen balloon, 0.80g of hydrogen is reacted with 6.4g of oxygen.

Write the balanced equation

2 H2 + O2 ---> 2 H2O

0.80 g H2 x (1 mole / 2g) = 0.4 moles H2

6.4 g O2 x (1 mole / 32g) = 0.2 moles O2

from balanced equation, 2 moles H2 react with 1 mole O2

therefore

0.4 moles H2 x (1 mole O2 / 2 moles H2) = 0.2 moles O2

since we need 0.2 moles O2 to react with all the H2 and we have exactly that amount, the amounts are said to be

"stoichiometric" and either reactant can be considered limiting

from the balanced equation, 2 moles H2 --> 2 moles H2O.. therefore

0.4 moles H2 x (2 moles H2O / 2 moles H2) = 0.4 moles H2O

Then

0.4 moles H2O x (18.0 g / mole) = 7.2 g H2O

8 0
3 years ago
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