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astraxan [27]
2 years ago
12

According to Image 2, which tectonic plate is composed of the largest percentage of oceanic lithosphere?

Chemistry
2 answers:
Luda [366]2 years ago
3 0
I think it would be African plate
VLD [36.1K]2 years ago
3 0

Answer:

I think the answer is A the Afracan plate

Explanation:

hope this helps:)

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What is the name of this compound? H single bonded to N with a pair of electron dots above and a single bond to H below, single
dybincka [34]

Answer:

Ethanamine (also known as ethylamine)

Explanation:

The compound that is requested by the question is ethanamine. Its trivial name is ethylamine.

It is a compound that contained the ethyl moiety (CH3CH2-) as well as the amine moiety (-NH2).

Ethanamine has a structure that can easily be determined by the statements in the question.

The structure of ethanamine is shown in the image attached.

4 0
3 years ago
The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make
natta225 [31]

Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

μ = \dfrac{1 \times 35}{1 + 35}

μ = \dfrac{35}{36}

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = \\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \  kg

μ  = 1.6139 × 10⁻²⁷ kg

r_o = 127 \ pm = 127*10^{-12} \ m

The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

where ;

J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8  \times  \pi ^2  \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2  } \times 3 ( 3 +1)

E_3= 2.5665 \times 10^{-21} \ J

We know that :

1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

8 0
2 years ago
A sample of 1.000 g of a compound containing carbon and hydrogen reacts with oxygen at elevated temperature to yield 0.692 g H₂O
ollegr [7]

Answer :

(a) 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) There is no other element present in the compound.

Explanation :

(a) Now we have to determine the masses of C and H in the sample.

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Mass of CO_2=3.381g

Mass of H_2O=0.692g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.381 g of carbon dioxide, \frac{12}{44}\times 3.381=0.922g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.692 g of water, \frac{2}{18}\times 0.692=0.0769g of hydrogen will be contained.

Thus, 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) Now we have to determine the compound contain any other elements or not.

Mass carbon + Mass of hydrogen = 0.922 g + 0.0769 g = 0.999 g ≈ 1 g

This means that there is no other element present in the compound.

3 0
3 years ago
What are the two processes underlying this image that turn peat into coal
Travka [436]

Answer:

Peatification and coalification

Explanation:

https://energyeducation.ca/encyclopedia/Coal_formation

8 0
3 years ago
Read 2 more answers
When the following equation is balanced, what is the coefficient of oxygen?
Gwar [14]

Answer:

The answer is 3

C2H5OH + O2 CO2 +H2O (unbalanced)

C2H5OH +3O2(g). 2CO2(g)+3H2O(balanced)

4 0
3 years ago
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