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bixtya [17]
3 years ago
7

Is a molecule the same as a compound ?

Chemistry
1 answer:
Alenkasestr [34]3 years ago
5 0
No a molecule is 2 different atoms bond
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Why do some elements form double and triple bonds during bonding
jekas [21]

Explanation:

Elements need a total of eight electrons to gain stability and look like a noble gas. So, they sometimes need sharing of two, four or even six electrons to complete their octate. So, they form double and triple covalent bonds. One more  the reason is the  interaction between the p orbitals of the combining atoms. for example  A double bond, as in ethene H2C=CH2, arises from one combination of the s orbitals and one combination of the p_y orbitals.

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Which type of fuel can only be used in certain geographic area? A. natural gas B. nuclear C. solar D. Biomass
poizon [28]

Answer:

A

Explanation:

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Can someone help me?
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Answer 19.9g. I’ve took the test last week at my uncle randy’s house
8 0
3 years ago
Which of the following compounds are soluble in aqueous solutions? Select ALL that are soluble.
sattari [20]
<h3>Answer:</h3>
  • K₂S
  • CaSO₄ (slightly soluble in water)
  • Pb(ClO₄)₂
<h3>Explanation:</h3>
  • Mg(OH)₂  and CoCO₃ are insoluble in water.
  • Solubility rules are used to determine whether a compound will be soluble or not soluble in aqueous form.
  • K₂S is soluble in aqueous solutions as all salts of Potassium are soluble in water.
  • Calcium (ii) sulfate is slightly soluble in water.
  • Hydroxides are insoluble in water except, KOH, NaOH and NH₄OH.and Barium hydroxide. Therefore, Mg(OH)₂ is an insoluble hydroxide.
  • CoCO₃ is insoluble since carbonates are insoluble except K₂CO₃, Na₂CO₃ and (NH₄)₂CO₃. .
  • Pb(ClO₄)₂ is soluble because all perchlorates are soluble.
7 0
3 years ago
When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
natka813 [3]

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

3 0
3 years ago
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