<span>Range = 88.5 Km/h - 94.5 Km/h</span><span>
</span>
Answer:
ΔH = 249 kJ/mol
Explanation:
The balanced reaction is:
2H₂O(g) → 2H₂(g) + O₂(g) (1)
To calculate the energy change to obtain one mole of H₂(g) from one mole of H₂O(g), the coefficients of the reaction (1) must be halved:
H₂O(g) → H₂(g) + 1/2O₂(g) (2)
The enthalpy of the reaction (2) is given by:
<em>Where 
: is the bond enthalpy of reactants and 
: is the bond enthalpy of products.</em>
<u>For the reactants we have the next bond energies:</u>
2 x (H-O) = 2 x (467)
<u>And the bond energies for the products are:</u>
H-H + (1/2) (O=O) = 436 + (1/2)(498)
So, the enthalpy of the reaction (2) is:
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Below are the choices that can be found from other sources:
A. The marlin has more momentum.
<span>B. The salmon has more momentum. </span>
<span>C. Both have the same momentum. </span>
<span>D. Both have no momentum.
</span>
Velocity is speed. Momentum is the product of velocity and mass. Because the Marlin has a much greater mass that means that it's momentum would also be greater.
<span>Lets look at this. momentum = mass x velocity </span>
<span>lets say that velocity for both fish is at 2 meters per second </span>
<span>Marlin : momentum = 500 x 2 = 1000 </span>
<span>Salmon : momentum = 20 x 2 = 40 </span>
<span>Therefore the answer is A</span>
It is about a billion times brighter.
<h3>What is the brightness of a star called?</h3>
- A star's apparent magnitude, or how bright it seems to be from Earth, and absolute magnitude, or how brilliant it looks at a standard distance of 32.6 light-years, or 10 parsecs, are words used by astronomers to describe stellar brightness.
<h3>How bright are planets compared to stars?</h3>
- Although planets and stars light up the night sky, planets usually appear much brighter than most stars.
- Astronomers use astronomical scales to measure the relative brightness of celestial bodies, and many planets fall within the range easily visible to the eye.
To learn more about planets from the given link
brainly.com/question/1286910
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Answer:
When x = 2.8 cm, 
When x = 5.5 cm, 
when x = 7.3 cm, 
When x = 11.0 cm, 
Explanation:
According to Biot-Savart law,
.......................(1)
R = 11.0 cm = 0.11 m
I = 17.0 A
N = 300 turns
When x₁ = 2.8 cm = 0.028 m
When x₂ = 5.5cm = 0.055 m
When x₃ = 7.3 cm = 0.073 m
When X₄ = 11.0 cm = 0.11 m
