Answer:
E = 5.69x10⁻²⁸m
Explanation:
To solve this question we neeed to convert the wavelength in meters to energy in joules using the equation:
E = hc / λ
<em>Where E is energy in joules, h is Planck's constant = 6.626x10⁻³⁴Js</em>
<em>c is light constant = 3.0x10⁸m/s</em>
<em>And λ is wavelength in meters = 349m</em>
Replacing:
E = 6.626x10⁻³⁴Js*3.0x10⁸m/s / 349m
E = 5.69x10⁻²⁸m
<h3>
Answer:</h3>
0.424 J/g °C
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Chemistry</u>
<u>Thermochemistry</u>
Specific Heat Formula: q = mcΔT
- q is heat (in Joules)
- m is mass (in grams)
- c is specific heat (in J/g °C)
- ΔT is change in temperature
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] m = 38.8 g
[Given] q = 181 J
[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C
[Solve] c
<u>Step 2: Solve for Specific Heat</u>
- Substitute in variables [Specific Heat Formula]: 181 J = (38.8 g)c(11.0 °C)
- Multiply: 181 J = (426.8 g °C)c
- [Division Property of Equality] Isolate <em>c</em>: 0.424086 J/g °C = c
- Rewrite: c = 0.424086 J/g °C
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.424086 J/g °C ≈ 0.424 J/g °C
Answer: C) the values of Kb and Kw
Explanation: i just took the test
Option 3 using forest land to build homes
C. both a and b
If a light bulb can last longer with the same amount of energy it is given, that means it can use less energy to do the same job compared to one that does not last longer with the same amount of energy it is given. It is much like how a more fuel efficient car will be able to go farther on the same tank of gas, but if you pair it with a car that doesn't have as great of an mpg, when they go the same distance, the car with the greater mpg spends less fuel.
If you don't have to use the energy when you aren't utilizing it, then you can conserve the energy for when you do need it.
