Erosion
waethring
deposition
pressure or heat
all of this can change the form of rock to another
Answer:
Correct: B
Explanation:
The first block wont move because:
Σf=3-3=0N.
The second block will move because:
Σf:5-3=2N (direction to the right).
The third block will move because:
Σf=3N (direction to the right).
The number of moles of silver oxide (I) needed to produce 4 moles of silver is 2 moles
<h3>Stoichiometry </h3>
From the question, we are to determine the number of moles of silver oxide (I) needed to produce 4 moles of silver
First, we will write the balaced chemical equation for the decomposition of silver oxide (I)
2Ag₂O(s) → 4Ag(s) + O₂(g)
This means, 2 moles of silver oxide (I) [Ag₂O] decomposes to give 4 moles of <u>silver </u>and 1 mole of oxygen gas.
From the <em>balanced chemical equation</em>, it is easy to deduce the number of moles of silver oxide (I) that would give 4 moles of silver.
Hence, the number of moles of silver oxide (I) needed to produce 4 moles of silver is 2 moles
Learn more on Stoichiometry here: brainly.com/question/18834543
Answer:
Explanation:
A <em>combustion reaction</em> is the reaction with oxygen along with the release of energy in form of heat or light.
Organic compounds (like CH₄) undergo combustion forming water and CO₂.
The combustion reaction of CH₄ is:
Hence, the first equation from the choices is not showing the combustion reaction of CH₄.
Not only organic compounds can undergo combustion. Metals and no metals can undergo combustion, i.e. metals and no metals can react with oxygen releasing light or heat.
The reaction of copper and oxygen (second choice) is a combustion reaction:
The formation of water (2H₂ + O₂ → 2H₂O) is other example of a combustion reaction where no organic compounds are involved.
On the other hand, the other two equations from the choice list are not reactions with oxygen, so they do not show combustion reactions.
Answer:
K = 361.369
Explanation:
C2H4(g) + H2O(g) → CH3CH2OH(g)
∴ ΔG°f(298.15K) CH3CH2OH(g) = - 174.8 KJ/mol
∴ ΔG°f(298.15) C2H4(g) = 68.4 KJ/mol
∴ ΔG°f(298.15) H2O(g) = - 228.6 KJ/mol
⇒ ΔG°f(298.15) = - 174.8 - ( - 228.6 + 68.4 ) = - 14.6 KJ/mol
∴ R = 8.314 E-3 KJ/mol.K
∴ T = 298.15 K
⇒ K = e∧(-(-14.6)/((8.314 E-3)(298.15)))
⇒ K = e∧(5.889)
⇒ K = 361.369