Answer:
compacting
Explanation:
i don't think there is very much explanation, the snow falls and compacts the ice to become giant lol
Either A)a color change has occurred or B)a smell has occurred
Answer:
Option (c)
Explanation:
Both the bullets have same acceleration because they both falls under the influence of acceleration due to gravity.
The bullet which is fired from the gun has some initial velocity but the bullet which is dropped has zero initial velocity.
the acceleration is acting on both the bullets which is equal to the acceleration due to gravity and they both in motion in the influence of gravity.
Answer:
The products of 1.0 kg of calcium phosphate reacting with 1.0 kg concentrated sulfuric acid (98% H2SO4 by mass) are:
- <em>1.3 kg of Calcium sulfate.</em>
- <em>0.6 kg of Phosphoric acid</em>
- <em>And 0.03 kg of unconsumed sulfuric acid and 0.02 kg of impurities in the sulfuric acid.</em>
Explanation:
Step 1.
The <em>balanced equation </em>of the reaction is required. For this case is:
Where:
- is Sulfuric Acid.
- is Calcium Phosphate.
- is Calcium Sulfate.
- is Phosphoric acid.
Step 2.
The <em>molar masses</em> (the weight of one mol) of each compound are required:
- Sulfuric Acid: 98.079 g/mol
- Calcium Phosphate: 310.18 g/mol
- Calcium Sulfate: 136.14 g/mol
- Phosphoric acid: 97.994 g/mol
Step 3.
The amount of moles of sulfuric acid and calcium phosphate avalaible are calculated using the corresponding <em>molar mass</em>.
- For sulfuric acid: (note that the real amount of acid is the purity times the weight)
- For calcium phosphate: .
Step 4.
From the balanced equation in Step 1. It is known that:
<em>3 moles of Sulfuric acid need 1 mol (no coefficient) of Calcium phosphate to react</em>.
Then is possible to determinate which of the reagents is first totally consumed (<em>limiting reagent</em>).
- For sulfuric acid: For 3 moles of sulfuric acid only one mol of calcium phosphate is required. Then the 9.99 moles would need 3.33 moles of calcium phosphate (more than available).
- For calcium phosphate: 1 mol of calcium phosphate will require 3 moles of sulfuric acid. Then, the 3.22 moles would need 9.67 moles of sulfuric acid (less than the 9.99 available).
Then, is possible to conclude that all the calcium phosphate would be consumed (It is the limiting reagent), and some sulfuric acid would remain.
Step 4.
Using the determined limiting reagent and, from the balanced equation of step 1, it is possible to know that:
<em>For each 1 mol of calcium phosphate consumed: 3 moles of calcium sulfate and, 2 moles of phosphoric acid would be produced.</em>
With this information:
- Calcium sulfate: produced.
- Phosphoric acid: produced.
- Sulfuric acid: As there is more sulfuric acid (9.99 mol) that the 9.67 required, then 0.32 mol remains after the calcium phosphate is depleted.
Step 5.
Using the molar masses in the step 2, is possible to calculate the weight of each species in the products.
- Calcium sulfate: produced.
- Phosphoric acid: produced.
- Sulfuric acid: unused and 0.02 kg of impurities in the original sulfuric acid (98% acid, and 2% impurities).
Answer:
After finding the electric potential VP at point P = Q/Чπϵ₀L ㏑(1+)
Explanation:
I believe it is a part C question.
The derivative of V and P will be directly proportional to the differential dq and the inverse of Чπϵ₀δ........
Please find detailed solution in the attached picture as i believe that is the answer to the part C question you are seeking for.