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2 years ago

• An element in group 2, period 4 has similar properties with

Chemistry

1 answer:

babymother [125]2 years ago

7 0

Answer: Elements in Group 2

Explanation: The periodic table was arranged by Dmitri Mendeleev specifically around similarites in their chemical behaviors. He found that as atomic number increases, at some point an element starts to react in a manner similar to a previous one. When that happened, he would place the larger element under the smaller one, and eventually noticed a periodicity in the table. Elements in a column (Groups) had similiar chemical properties. We know today that these similarities are due to the electron configuration, and that these configurations repeat themselves. He left gaps in the table when he could find an existing element with properties similar to others in that group. I big leap of faith, but it worked. Elements for those missing boxes were eventually discovered.

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Perform the following operation

Paul [167]

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \:2.8 × {10}^{-5}$

____________________________________

$\large \tt Solution \: :$

$\qquad \tt \rightarrow \: \dfrac{2.36 \times 10 {}^{ - 2} }{8.4 \times {10}^{2} }$

$\qquad \tt \rightarrow \: \dfrac{2.36 }{8.4 } \times \cfrac{10 {}^{ - 2} }{10 {}^{2} }$

$\qquad \tt \rightarrow \: 0.28 \times 10 {}^{ - 2 - 2}$

$\qquad \tt \rightarrow \: 0.28 \times 10 {}^{ - 4}$

$\qquad \tt \rightarrow \: 2.8 \times 10 {}^{ - 5}$

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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1 year ago

C3H8+3O2 = 3CO2+4H2O what is the enthalpy combustion

dezoksy [38]

Answer:

$\Delta H_{comb}=2043.85kJ/mol$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it possible for us to set up the expression for the calculation of the enthalpy change as shown below:

$\Delta H_r=-\Delta H_{comb}=3\Delta _fH_{CO_2}+4\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-3\Delta _fH_{O_2}$

Thus, given the values of the enthalpies of formation on the attached file, we obtain: $-\Delta H_{comb}=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-103.85kJ/mol)-3(0kJ/mol)\\\\-\Delta H_{comb}=-2043.85kJ/mol\\\\\Delta H_{comb}=2043.85kJ/mol$

Best regards!

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2 years ago

How does the air on a hot day compare with the air on a cold day?

MrMuchimi

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wariber [46]

Answer:

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3 years ago

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andreev551 [17]

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