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3 years ago

What makes this inauguration different from the past?

Chemistry

1 answer:

Sveta_85 [38]3 years ago

7 0

Answer:

there's just a lot of drama around grumpy trumpy's StOlEN eLecTiOn which are all just fake claims. all his hardcore followers are brainwashed into believing anything he says, so people (especially people who are inside the capitol) fear that trump's followers might incite violence AGAIN even after they broke into the Capitol on the 6th.

That's why they have the National Guard and thousands of people to keep the building secure during Biden's Inauguration.

hope this helps!

You might be interested in

The photon of light that is emitted as an electron drops back to its original orbit is:

Alex73 [517]

Answer:

energy

Explanation:

The photon of light that is emitted as an electron drops back to its original orbit is energy and this energy is released during de-excitation process.

The electron is jumped into higher level and back into lower level by absorbing and releasing the energy.

The process is called excitation and de-excitation.

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits. For example if electron jumped from K to L it must absorbed the energy which is equal the energy difference of these two level. The excited electron thus move back to lower energy level which is K by releasing the energy because electron can not stay longer in higher energy level and comes to ground state.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum

3 0

2 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

Mendeleev put elements in groups while creating the periodic table. What important role do groups play in the structure of the p

NikAS [45]

Answer:

Groups show similar chemical properties about the elements

Explanation:

The periodic table is divided into groups and periods. Groups are vertical while periods are horizontal.

All elements in the same group possess the same number of outermost shell electrons. This number of outermost shell electrons controls the reactivity of the elements in the group.

Hence, groups show similar chemical properties of the elements in them, Groups separate elements into families which assist in the speedy study of elements.

3 0

3 years ago

Suppose that in an ionic compound, "m" represents a metal that could form more than one type of ion. in the formula mf2 , the ch

geniusboy [140]

F (Fluorine) is in column (group/family) VIIA, or the "halogens". When you see the halogens (Fluorine, Chlorine, Bromine, and Iodine) in combination with a metal, each halogen atom present will carry a -1 charge. We can see that the atom has no charge, so the metal must cancel out the negative charges brought by the two fluorine atoms.
(Charge on m) + 2*(charge on fluorine) = 0
(Charge on m) + 2*(-1) = 0
(Charge on m) - 2 = 0
Charge on m ion = +2

3 0

3 years ago

Which of the following reactions could be an elementary reaction? 2 NO2(g) + F2(g) → 2NO2F(g) Rate = k[NO2][F2] H2(g) + Br2(g) →

podryga [215]

Answer: The correct answer is $NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2]$

Explanation:

Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.

For the given reactions:

Equation 1: $2NO_2(g)+F_2(g)\rightarrow 2NO_2F(g);Rate=k[NO_2][F_2]$

Molecularity of the reaction = 2 + 1 = 3

Order of the reaction = 1 + 1 = 2

This is not considered as an elementary reaction.

Equation 2: $H_2(g)+Br_2(g)\rightarrow 2HBr(g);Rate=k[H_2][Br_2]^{1/2}$

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = $1+\frac{1}{2}=\frac{3}{2}$

This is not considered as an elementary reaction.

Equation 3: $NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2]$

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = 1 + 1 = 2

This is considered as an elementary reaction.

Equation 4: $NO_2(g)+CO(g)\rightarrow NO(g)+CO_2(g);Rate=k[NO_2]^2$

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = 2 + 0 = 2

In this equation, the order with respect to each reactant is not equal to its stoichiometric coefficient which is represented in the balanced chemical reaction. Hence, this is not considered as an elementary reaction.

Hence, the correct answer is $NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2]$

3 0

3 years ago