The change in the velocity = 4 m/s
Acceleration = 4 m/s²
<h3>Further explanation</h3>
Given
vo = initial velocity = 4 m/s
vf = final velocity = 8 m/s
t = 1 s
Required
The change in the velocity
Acceleration
Solution
the change in velocity =
Acceleration = ratio of a change in velocity and the time
Input the value :
Reducing acidity of chyme :Acidic chyme entering the duodenum stimulates the release of secretin from the small intestinal glands.
Answer:
A 50-mL volumetric cylinder with 0.1-mL accuracy scale should be used for this purpose since three significant figures of accuracy are required.
Explanation:
Hello,
A 50-mL volumetric cylinder with 0.1-mL accuracy scale should be used for this purpose since three significant figures of accuracy are required.
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Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
1. D
2. DR
3. SR
4. DR
5. C
6. SR
7. S
8. D
10. C
