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3 years ago

8

As we move from left to right across the periodic table, what is the general trend? A) Atomic radii increase. B) Electronegaviti

y decreases. C) Nuclear shielding increases. D) Metallic character decreases.

2 answers:

inysia [295]3 years ago

5 0

Answer:

So it was D?

Thanks

Explanation:

elena55 [62]3 years ago

4 0

B. Electronegavitiy decreases

Electron affinity decreases as we proceed down a group.

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MgO + H2O = Mg(OH)2 is an example of a _____ chemical reaction.

algol [13]

a) synthesis- or combination- reaction

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3 years ago

Which of the following reactions is a synthesis reaction?

Vedmedyk [2.9K]

<h3>Answer:</h3>

B) 4H₂(g) + O₂(g) ⟶ 2H₂O(l)

<h3>Explanation:</h3>

Chemical reactions occur when compounds or elements combine to form new compounds or other elements.
Chemical reactions may be classified into various types which include synthesis reactions, replacement reaction, decomposition reactions, and precipitation reactions among others.
In our case, we were supposed to identify a synthesis reaction.
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2 years ago

At chemical equilibrium the amount of blank because

Norma-Jean [14]

Answer:

In a chemical reaction, chemical equilibrium is the state in which the forward reaction rate and the reverse reaction rate are equal. The result of this equilibrium is that the concentrations of the reactants and the products do not change.

Explanation:

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2 years ago

4) list 3 protest rights you have that are protected under the law. How can you protect these rights? George Floyd’s ?

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Peaceful Protest, Signing Pettitons, contacting mayors/governors and writing emails/sending letters

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2 years ago

. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r

Tom [10]

Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

Explanation : Given,

Wavelength = $435.8nm=435.8\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $435.8\times 10^{-9}m$

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}$

$E=4.56\times 10^{-19}J$

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

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3 years ago