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hichkok12 [17]
3 years ago
15

Which statement is true about trends shown on the electromagnetic scale.

Chemistry
2 answers:
lawyer [7]3 years ago
6 0
Remember that <span>λ, or wavelength, is proportional to (1/f), where f=frequency. That means that wavelength and frequency are inversely related. When one goes up, the other goes down, and vice versa. That means answers C and D are incorrect because those answers have wavelength and frequency in a direct relationship.

Radiant energy is directly proportional to frequency. When frequency increases, radiant energy increases. That means A) </span><span>As wavelength decreases, radiant energy and frequency increase is the correct answer.</span>
kicyunya [14]3 years ago
4 0

Answer:

A: As wavelength decreases, radiant energy and frequency increse

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Which polynomials, given in square inches, could represent the area of a square with whole number side lengths if x is a whole n
erica [24]

<u>Answer:</u> The correct answers are x^2-9\text{ and }x^2-100

<u>Explanation:</u>

To find the polynomials which could represent the are of a square having side x greater than 2, we need to find the value of 'x' for all the given polynomials.

From the given options:

  • 1. x^2-9

x^2=9\\x=\sqrt{9}\\x=3,-3

x = -3 is ignored.

  • 2. x^2-100

x^2=100\\x=\sqrt{100}\\x=\pm10\\x=10,-10

x = -10 is ignored

  • 3. x^2-4x+4

To solve this we use the quadratic formula:

\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Putting values of a, b and c, we get:

x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(4)}}{2\times 1}\\x=2,2

As, x comes out to be 2 and is not greater than 2. Hence, this is not considered.

  • 4. x^2+10x+25

Solving for 'x' by splitting the middle term:

\Rightarrow x^2+5x+5x+25\\\Rightarrow x(x+5)+5(x+5)\\x=-5,-5

Hence, this is ignored.

  • 5. x^2+15x+36

Solving for 'x' by splitting the middle term:

\Rightarrow x^2+12x+3x+36\\\Rightarrow x(x+12)+3(x+12)\\x=-12,-3

Hence, this is ignored.

So, the correct polynomials are x^2-9\text{ and }x^2-100

3 0
3 years ago
Read 2 more answers
The specific heat of water is 4.184Jg ∘C. Determine the final temperature when 600.0 g water at 75.5∘C absorbs 5.90×104 J of ene
sesenic [268]

Answer:

T_2=98.5^{\circ}

Explanation:

Given that,

The specific heat of water is 4.184Jg°C

Mass, m = 600 g

Initial temperature, T₁ = 75.5°C

We need to find the final temperature. We know that heat absorbed is given by :

Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\\\T_2=\dfrac{Q}{mc}+T_1\\\\T_2=\dfrac{5.9\times 10^4}{600\times 4.184}+75\\\\T_2=98.5^{\circ}

So, the final temperature is equal to 98.5^{\circ}.

3 0
3 years ago
What is the concentration of a solution?
Anon25 [30]

Concentration Definition. In chemistry, concentration refers to the amount of a substance per defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution. Concentration usually is expressed in terms of mass per unit volume.

7 0
3 years ago
During the light-independent reaction, carbon dioxide is fixed by adding it to a A. 2-carbon compound B. 3-carbon compound C. 4-
Scrat [10]
During the light independent reaction, carbon dioxide is fixed by adding it to a <span>5-carbon compound</span>
8 0
4 years ago
Read 2 more answers
Calculation of ph from hydrogen ion concentration what is the ph of a solution that has an h1 concentration of (a) 1.75 3 1025 m
faltersainse [42]
<h2>Solutions:</h2>

<u>Case a:</u> Finding pH for [H⁺] = 1.75 × 10⁻⁵ mol/L :

As we know pH is given as,

                                  pH  = -log [H⁺]

Putting value,

                                  pH  = -log [1.75 x 10⁻⁵]

                                 pH =  4.75

<u>Case b:</u> Finding pH for [H⁺] = 6.50 × 10⁻¹⁰ mol/L :

As we know pH is given as,

                                  pH  = -log [H⁺]

Putting value,

                                  pH  = -log [6.50 × 10⁻¹⁰]

                                 pH =  9.18

<u>Case c:</u> Finding pH for [H⁺] = 1.0 × 10⁻⁴ mol/L :

As we know pH is given as,

                                  pH  = -log [H⁺]

Putting value,

                                  pH  = -log [1.0 × 10⁻⁴]

                                 pH =  4

<u>Case d:</u> Finding pH for [H⁺] = 1.50 × 10⁻⁵ mol/L :

As we know pH is given as,

                                  pH  = -log [H⁺]

Putting value,

                                  pH  = -log [1.50 × 10⁻⁵]

                                 pH =  4.82

7 0
3 years ago
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