When encountering another vessel in darkness, the visible white, red and green lights indicate you are approaching a powerboat head-on.
In this case neither vessels has the right of way. Both boat operators must take early action to stay out of the way for other vessels. The operators must reduce their speed and steer to starboard.
<h2>Further Explanation</h2>
The visible white, red and green lights are all navigation lights with different purposes. However, the four common navigational lights are sidelights, sternlight, Masthead light and all-round white light.
Sidelights are also referred to as combination lights. It is the combination of red and green light and visible to another vessel that approaching head on or from the side. The green light is the boat starboard (right hand side) and the red light is the port side (left) of a vessel.
The sternlight is the white light seen from behind the vessel.
The masthead light is a white light. The white light shines forward to all sides of the vessels. It is on all power driven vessels.
All-round white light is a white light used to combine sternlight and masthead light to form a single white light so that it can boldly be seen by other vessels.
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KEYWORDS:
- boat
- head-on
- masthead light
- white light
- operators
Answer:
b. 4.4
Explanation:
pH is related to the concentration of H₃O⁺ through the following equation:
pH = -log([H₃O⁺]) = -log(4 x 10⁻⁵)
pH = 4.4
A gamma ray has a mass and atomic number of 0 meaning it doesn’t effect the nucleus.
E.g
60 60 0
Co —> Co + gamma ray
27 27 0
Answer:
The answer to your question is below
Explanation:
Data
Volume = 1000 ml
Concentration = 2M
molecule = NaCl
Process
1.- Calculate the number of moles of NaCl
Molarity = moles/Volume
-Solve for volume
moles = Molarity x Volume (liters)
-Substitution
moles = 2 x 1
-Result
moles = 2
2.- Determine the molar mass of NaCl
NaCl = 23 + 35.5 = 58.5 g
3.- Calculate the mass of NaCl to prepare the solution
58.5 g ----------------- 1mol
x ----------------- 2 moles
x = (2 x 58.5) / 1
x = 117g
4.- Weight 117 g of NaCl, place them in a volumetric flask (1 l), and add enough water to prepare the solution.
Answer:
A. DH° = –36 kJ
Explanation:
It is possible to obtain DH° of a reaction by the sum of DH° of half reactions. The DH° of the reaction:
B₂H₆(g) → 2B(s) + 3H₂(g)
Could be obtained from:
<em>(1) </em>2B(s) + 1.5O₂(g) → B₂O₃(s) DH° = –1273kJ
<em>(2) </em>B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g) DH° = –2035kJ
<em>(3) </em>H₂(g) + 0.5O₂(g) → H₂O(g) DH° = –242kJ
The sum of (2) - (1) gives:
B₂H₆(g) + 1.5O₂(g) → 2B(s) + 3H₂O(g) DH° = -2035kJ - (-1273kJ) = -762kJ
Now, this reaction - 3×(3):
B₂H₆(g) → 2B(s) + 3H₂(g) DH° = -762kJ - (3×-242kJ) = -36kJ
Thus, right answer is:
<em>A. DH° = –36 kJ</em>
