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ki77a [65]
3 years ago
14

What is the process all rocks are continuously in during which that can change into other types of rocks

Chemistry
1 answer:
andrezito [222]3 years ago
6 0
The rock cycle

The rock cycle can have sedimentary, igneous, or metamorphic rocks going back and forth changing into different rocks.
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Nitric acid is a strong acid, sodium hydroxide is a strong base, and sodium nitrate is a soluble salt. Which of the following is
Evgesh-ka [11]
Can’t help with this but good luck
4 0
3 years ago
A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains
Taya2010 [7]

Answer:

[Cu²⁺] = 2.01x10⁻²⁶

Explanation:

The equilibrium of Cu(CN)₄²⁻ is:

Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻

And Kf is defined as:

Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴

As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:

[Cu²⁺] = 0

[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M

[Cu(CN)₄²⁻] = 2.2x10⁻³

Some [Cu²⁺] will be formed and equilibrium concentrations will be:

[Cu²⁺] = X

[CN⁻] = 0.3212M + 4X

[Cu(CN)₄²⁻] = 2.2x10⁻³ - X

<em>Where X is reaction coordinate</em>

<em />

Replacing in Kf equation:

1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴

1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0

Solving for X:

X = 2.01x10⁻²⁶

As

[Cu²⁺] = X

<h3>[Cu²⁺] = 2.01x10⁻²⁶</h3>

5 0
3 years ago
Isotope Atomic Mass (amu) Percent Abundance
exis [7]

Answer:

The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.

Explanation:

Given data:

Atomic mass of silicon= ?

Percent abundance of Si-28 = 92.21%

Atomic mass of Si-28 = 27.98 amu

Percent abundance of Si-29 = 4.70%

Atomic mass of  Si-29 = 28.98 amu

Percent abundance of Si-30 = 3.09%

Atomic mass of  Si-30 = 29.97 amu

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass = (92.21×27.98)+(4.70×28.98)+(3.09×29.97) /100

Average atomic mass =  2580.04 +136.21+92.61 / 100

Average atomic mass = 2808.86 / 100

Average atomic mass  = 28.08amu.

The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.

7 0
3 years ago
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Which of the following structures surround cells?
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Cells are surrounded by #2 , Cell membrane and Cell wall thats what protects the cell
4 0
3 years ago
BONUS QUESTION - It takes 194 mL of a 0.5 M NaOH solution to neutralize 300
Strike441 [17]

Answer:

1.7 mol/dm3

Explanation:

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6 0
3 years ago
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