3 years ago

Help me pls rawr uwu SSUsSY

Chemistry

2 answers:

cricket20 [7]3 years ago

7 0

Help 2 what the answer for this?

MAXImum [283]3 years ago

7 0

It would be A divided

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We have a photon with the wavelength of 605 nm. What is the frequency of this<br> photon?

Kipish [7]

Answer:

Explanation:

Speed of light = Wavelength x Frequency.

605 nm x Frequency = 3x10^8 m/s

6.05 x 10^-7 m x Frequency = 3x10^8 m/s

Frequency = 4.96 x 10^14 Hz

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3 years ago

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Write the condensed electron configurations for the following atoms, using the appropriate noble-gas core abbreviations.

MatroZZZ [7]

Answer:

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^4

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3 years ago

The way light bounces off a minerals surface is describe by the minerals

kirill [66]

Would love to answer but "descibe by the minerals" makes no sense

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3 years ago

A sample weighing 3.110 g is a mixture of Fe 2 O 3 (molar mass = 159.69 g/mol) and Al 2 O 3 (molar mass = 101.96 g/mol). When he

grandymaker [24]

Answer:

The mass fraction of ferric oxide in the original sample : $\frac{723}{3110}$

Explanation:

Mass of the mixture = 3.110 g

Mass of $Fe_2O_3=x$

Mass of $Al_2O_3=y$

After heating the mixture it allowed to react with hydrogen gas in which all the ferric oxide reacted to form metallic iron and water vapors where as aluminum oxide did not react.

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

Mass of mixture left after all the ferric oxide has reacted = 2.387 g

Mass of mixture left after all the ferric oxide has reacted = y

$x=3.110 g- y=3.110 g - 2.387 g = 0.723 g$

The mass fraction of ferric oxide in the original sample :

$\frac{0.723 g}{3.110 g}=\frac{723}{3110}$

5 0

3 years ago

Enter the Ksp expression for the solid AB2 in terms of the molar solubility x. AB2 has a molar solubility of 3.72×10−4 M. What i

AlekseyPX

Answer:

2.06 × 10⁻¹⁰

Explanation:

Let's consider the solution of a generic compound AB₂.

AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

We can relate the molar solubility (S) with the solubility product constant (Kps) using an ICE chart.

AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

I 0 0

C +S +2S

E S 2S

The solubility product constant is:

Kps = [A²⁺] × [B⁻]² = S × (2S)² = 4 × S³ = 4 × (3.72 × 10⁻⁴)³ = 2.06 × 10⁻¹⁰

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3 years ago