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Gelneren [198K]
3 years ago
7

What happens when molten material is heated from Earth’s core?

Chemistry
1 answer:
shusha [124]3 years ago
4 0
B It rises up towards the crust
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If the pressure of a sample of gas is doubled at constant temperature, what happens to the volume of the gas?
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The volume gas will stay constant 
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(ii) The Cu2+ ions are attracted to the negative electrode, where they are reduced to produce copper atoms.
allochka39001 [22]
Electrolyte for the electrolysis is copper(II) sulfate
CuSO₄ = Cu²⁺ + SO₄²⁻

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(anode +) 2H₂O - 4e⁻ = O₂ + 4H⁺
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Which of the following statements accurately describes enzymes?
aniked [119]
The answer is a. temperature and pH can affect how the enzymes work
5 0
2 years ago
So2−3(aq)→so2−4(aq) (basic solution) express your answer as a chemical equation. identify all of the phases in your answer. so32
alexira [117]

Answer:

SO_{3}^{2-}(aq.)+2OH^{-}(aq.)\rightarrow SO_{4}^{2-}(aq.)+H_{2}O(l)+2e^{-}

Explanation:

The given reaction is an example of oxidation reaction. The step-by-step balancing procedure has been shown below:

Chemical equation: SO_{3}^{2-}(aq.)\rightarrow SO_{4}^{2-}(aq.)

Balance H and O in basic medium: SO_{3}^{2-}(aq.)+2OH^{-}(aq.)\rightarrow SO_{4}^{2-}(aq.)+H_{2}O(l)

Balance charge: SO_{3}^{2-}(aq.)+2OH^{-}(aq.)\rightarrow SO_{4}^{2-}(aq.)+H_{2}O(l)+2e^{-}

Balanced chemical equation:

SO_{3}^{2-}(aq.)+2OH^{-}(aq.)\rightarrow SO_{4}^{2-}(aq.)+H_{2}O(l)+2e^{-}

6 0
3 years ago
If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
Semenov [28]

Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

<em>SO2Cl2 -----> SO2(g) + Cl2(g) </em>

<em>Kc= 2.99 x 10^-7 at 227 degrees celcius </em>

<em>If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?</em>

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 ---------> SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

<em>[Cl₂] = 2.24x10⁻⁴ M</em>

5 0
3 years ago
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