The reason why a wave get taller as it gets closer to shore is that the shallow water at the bottom of the wave makes the wave length smaller. Option B
<h3>What is a wave?</h3>
A wave is a disturbance that occurs along a medium which transmits energy. Now we now that waves travel from place to place. The horizontal distance that is travelled by a wave is what we call the wavelength of the wave.
As the wave approaches the shore, the wave tends to slow down because it is dragged from beneath. In the process, the wave grows taller. Thus, the reason why a wave get taller as it gets closer to shore is that the shallow water at the bottom of the wave makes the wave length smaller. Option B
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The correct answer is A. O^2-
<span>When a chemical reaction occurs and the thermometer in the container records a drop in temperature t</span>he reaction is exothermic because heat was released by the reaction.
<span>B. The reaction is exothermic because heat was released by the reaction. is your answer.
Hoped I helped!</span>
This is an incomplete question, here is a complete question.
The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?
Answer : The time taken will be, 17.0 hr
Explanation :
Expression for rate law for first order kinetics is given by:

where,
k = rate constant = 
t = time passed by the sample = ?
a = initial concentration of the reactant = 0.080 M
a - x = concentration left = 0.053 M
Now put all the given values in above equation, we get


Therefore, the time taken will be, 17.0 hr
Answer:
- The molarity of the student's sodium hydroxide solution is 0.0219 M
Explanation:
<u>1) Chemical reaction.</u>
a) Kind of reaction: neutralization
b) General form: acid + base → salt + water
c) Word equation:
- sodium hydroxide + oxalic acid → sodium oxalate + water
d) Chemical equation:
- NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O
b) Balanced chemical equation:
- 2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O
<u>2) Mole ratio</u>
- 2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O
<u>3) Starting amount of oxalic acid</u>
- mass = 28 mg = 0.028 g
- molar mass = 90.03 g/mol
- Convert mass in grams to number of moles, n:
n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol
<u>4) Titration</u>
- Volume of base: 28.4 mL = 0.0248 liter
- Concentration of base: x (unknwon)
- Number of moles of acid: 2.52 mol (calculated above)
- Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)

That means that there are 0.000622 moles of NaOH (solute)
<u>5) Molarity of NaOH solution</u>
- M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M
That is the correct number using <em>three signficant figures</em>, such as the starting data are reported.