Answer:
Explanation:
Let's consider the following chemical equilibrium:
CaCO₃(s) ⇄ CaO(s) + CO₂(g)
Given the pressure equilibrium constant Kp = pCO₂
We can calculate the concentration equilibrium constant (Kc) using the following expression.
where,
R is the ideal gas constant
T is the absolute temperature
Δn(g) = moles of gaseous products - moles of gaseous reactants = 1 - 0 = 1
The expression for this reaction is:
Magnesium ribbon dissolves in zinc nitrate and copper nitrate solutions.
Even though the question is incomplete and refers to your practical work, however, I will try to help you as much as I can.
Metals dissolves in solutions of other metals that are lower than them in the electrochemical series. Hence, copper strip will show no change in magnesium nitrate or zinc nitrate solution. A zinc strip will not show any change in magnesium nitrate or zinc nitrate.
However, a magnesium ribbon will dissolve very quickly in zinc nitrate and copper nitrate solutions.
Learn more: brainly.com/question/14396802
a. malleable, flexible, and ductile
Answer:
Asteroids are made of rock, metals, and other elements. Some even contain water. Asteroids that are mostly stone sometimes are more likely to lose piles of rubble. Asteroids that are mostly iron are more rock-solid.
Explanation:
Answer:
0.0876L of 0.478M Na₃PO₄ are needed
Explanation:
The reaction of calcium chloride, CaCl₂, with sodium phosphate, Na₃PO₄ is:
3CaCl₂ + 2Na₃PO₄ → Ca₃(PO₄)₂ + 6NaCl
<em>Where 3 moles of calcium chloride react with 2 moles of sodium phosphate to produce 1 mole of calcium phosphate.</em>
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To solve this question we need to find moles of CaCl₂ added. Using the reaction we can find the moles of Na₃PO₄ that are needed to react completely and the volume using its concentration:
<em>Moles CaCl₂:</em>
0.225L * (0.279mol / L) = 0.0628moles of CaCl₂
<em>Moles Na₃PO₄:</em>
0.0628moles of CaCl₂ * (2mol Na₃PO₄ / 3 mol CaCl₂) =
0.0419moles of Na₃PO₄
<em>Volume 0.478M Na₃PO₄:</em>
0.0419moles of Na₃PO₄ * (1L / 0.478mol) =
<h3>0.0876L of 0.478M Na₃PO₄ are needed</h3>
