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2 years ago

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Chemistry

1 answer:

Luba_88 [7]2 years ago

6 0

Answer: The weight of the object on earth =441N

Weight of the object on moon = 72N

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What is the energy of a photon of yellow light with a frequency of 5.2 x 10¹⁴ Hz?<br><br>

3241004551 [841]

Answer:

3.4x10⁻¹⁹J is the energy of the photon

Explanation:

To find the energy of a photon from its frequency we must use the equation:

E = hf

<em>Where E is the energy of the photon in J, our incognite</em>

<em>h is Planck's constant = 6.6262x10⁻³⁴Js</em>

<em>And f is the frequency = 5.2x10¹⁴Hz = 5.2x10¹⁴s⁻¹</em>

<em />

Replacing:

E = 6.6262x10⁻³⁴Js*5.2x10¹⁴s⁻¹

E = 3.4x10⁻¹⁹J is the energy of the photon

<em />

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2 years ago

Oxidation of Nitrogen in N2O3

kirill [66]

Explanation:

<h3>oxidation of Nitrogen in N2O3 is </h3><h2>+3</h2>

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1 year ago

What is the weight of a 30.0 kg object?<br> 3.0 N<br> 300 N<br> 6.8 N<br> 132 N

andreyandreev [35.5K]

Explanation:

$w \: = mg \\ = 30\:kg \times 10\:ms^{-2}\\=300\:kg\:ms^{-2} \\ = 300\:N$

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3 years ago

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If all three metals were exposed to extremely high temperatures, which would be the last to melt?

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Copper would be the last to melt because it has the highest melting point!

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2 years ago

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Write the rates for the following reactions in terms of the disappearance of reactants and appearance of products: (a) 302 .....

Oduvanchick [21]

Answer:

Explanation:

$\mathbf{From \ the \ information \ given:} \\ \\ \mathbf{The \ rates \ of \ the \ f ollowing \ reactions \ can \ be \ expressed \ as \ follows:}$

(a)

$\mathbf{3O_2 \to 2O_3} \\ \\ \\ \mathbf{-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{2}{3} \dfrac{d[O_3]}{dt}}$

(b)

$\mathbf{C_2H_6 \to C_2H_4 + H_2} \\ \\ \\ \mathbf{ -\dfrac{d[C_2H_6]}{dt}= \dfrac{d[C_2H_4]}{dt}=\dfrac{d[H_2]}{dt}}$

(c)

$\mathbf{ClO^-+Br^- \to BrO^-+Cl^-} \\ \\ \\ \mathbf{ -\dfrac{d[ClO^-]}{dt}= -\dfrac{d[Br^-]}{dt} = \dfrac{d[BrO^-]}{dt} = \dfrac{d[Cl^-]}{dt} }$

(d)

$\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\ \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}$

(e)

$\mathbf{2AsH_3 \to 2As + 3H_2} \\ \\ \\ \mathbf{-\dfrac{1}{2}\dfrac{d[AsH_3]}{dt}=\dfrac{1}{2}\dfrac{d[As]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}}$

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2 years ago