Answer:
During photosynthesis, plants take in carbon dioxide (CO2) and water (H2O) from the air and soil. This transforms the water into oxygen and the carbon dioxide into glucose. The plant then releases the oxygen back into the air, and stores energy within the glucose molecules.
Explanation:
Brainliest please
Answer:
The particles in a solid vibrate in place.
Explanation:
The particles in a solid are close and tightly packed together so they have no room to flow past each other like the particles in a liquid can.
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Answer:
a) Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) 3.14g must be added
Explanation:
a) For the reaction:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)
As you see, there are 8 moles of water in reactants and 2 moles of oxygen in octahydrate, thus, water moles must be 10:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + NH₃(g)
To balance hydrogens, the other coefficients are:
Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:
6.5 g × (1mol / 315.48g) =<em> 0.0206moles of Ba(OH)₂.8H₂O</em>. Thus, moles of NH₄SCN that must be used for a complete reaction are:
0.0206moles of Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = <em>0.0412moles of NH₄SCN</em>. In grams:
0.0412moles of NH₄SCN × ( 76.12g / 1mol) = <em>3.14g must be added</em>
Answer:
a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.
Explanation:
The change in free energy (ΔG) that is, the <u>energy available to do work</u>, of a system for a constant-temperature process is:

-
When ΔG < 0 the reaction is spontaneous in the forward direction.
- When ΔG > 0 the reaction is nonspontaneous. The reaction is
spontaneous in the opposite direction.
- When ΔG = 0 the system is at equilibrium.
If <u>both ΔH and ΔS are positive</u>, then ΔG will be negative only when the TΔS term is greater in magnitude than ΔH. This condition is met when T is large.