Answer:
6.5 meters
Explanation:
Background Knowledge:
The most familiar example of breaking wave is the breaking of water surface waves on a coastline.
Explanation:
The process of Wave breaking in most of the cases occurs when the amplitude reaches the point that the crest of the wave actually overturns
Factors by which waves are produced:
Different type of waves are produced and varied according to different type of factors. These factors are
Wind direction
Type of swell
Sea floor features
Slope of sea bed
Direct answer of the question:
In general rule, Waves start to break on reaching a water depth of 1.3 times the wave height.
In this question wave height is 5 meters so wave breaks on
5 × 1.3 = 6.5 meters
The masses can be found by substractions:
- Mass of CaSO₄.H2O (hydrate):
16.05 g - 13.56 g = 2.49 g
15.07 g - 13.56 g = 1.51 g
- The mass of water is equal to the difference between the mass of the hydrate and the mass of the anhydrate:
2.49 g - 1.51 g = 0.98 g
- The percent of water is found by the formula:
massWater ÷ massHydrate * 100%
0.98 g ÷ 2.49 g * 100% = 39.36%
- The mole of water is calculated using water's molecular weight (18g/mol):
0.98 g ÷ 18 g/mol = 0.054 mol water
- A similar procedure is made for the mole of salt (CaSO₄ = 136.14 g/mol)
1.51 g ÷ 136.14 g/mol = 0.011 mol CaSO₄
- The ratio of mole of water to mole of anhydrate is:
0.054 mol water / 0.011 mol CaSO₄ = 0.49
In other words the molecular formula for the hydrate salt is CaSO₄·0.5H₂O
If the cyclist rode at an average speed of 10mph for 15 miles...
we can solve by dividing the distance by the speed to get time using the equation...
Δspeed = Δdistance / Δtime
Δtime = Δdistance / Δspeed
Δtime = 15 miles / 10 mph = 1.5 hours
A second order reaction varies with the square of the concentration of the reactant. Therefore, halving the concentration will reduce the rate of reaction by a factor of 4.
The answer is E.
Answer is: sodium (Na) and iodine (I₂).
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First ionic bonds in this salt are separeted
because of heat:
</span>NaI(l) → Na⁺(l) + I⁻(l).
Reaction of reduction
at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.
2Na⁺(l) + 2e⁻ → 2Na(l).
Reaction of oxidation
at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.
The anode is positive
and the cathode is negative.
