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lidiya [134]
4 years ago
9

A news report is an example of what kind of literature?

Chemistry
2 answers:
dangina [55]4 years ago
7 0

Answer:

1) I dont knoe

2) Poetry

Explanation:

poizon [28]4 years ago
7 0

Answer: Symbolic language is used by an author to hint at deeper meanings.

Explanation: penn foster

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In which type of reaction do two or more substances combine to produce a single substance?
melomori [17]
The correct answer is option 1. Synthesis reaction is the chemical reaction which involves two or more substances combining to form a single substance. An example is burning magnesium. The balanced reaction is 2Mg + O2 = 2MgO.
7 0
3 years ago
Read 2 more answers
What is the molar mass of 1-butene if 5.38 × 1016 molecules of 1-butene weigh 5.00 μg?
xenn [34]
<h3>Answer:</h3>

             Molar Mass =  56 g.mol⁻¹

<h3>Explanation:</h3>

Data Given:

                   Mass  =  5.00 μg  =  5.0 × 10⁻⁶ g

                   Number of Molecules  =  5.38 × 10¹⁶ Molecules

Step 1: Calculate Moles of 1-Butene:

                      As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Particles and Avogadro's Number is given as,

                          Number of Moles  =  Number of Particles ÷ 6.022 × 10²³

Putting values,

                          Number of Moles  =  5.38 × 10¹⁶ Molecules ÷ 6.022 × 10²³

                          Number of Moles  =  8.93 × 10⁻⁸ Moles

Step 2: Calculate Molar Mass of 1-Butene:

As,

                          Mole  =  Mass ÷ M.Mass

Solving for M.Mass,

                          M.Mass  =  Mass ÷ Mole

Putting values,

                          M.Mass  =  5.0 × 10⁻⁶ g ÷ 8.93 × 10⁻⁸ mol

                          M.Mass =  55.99 g.mol⁻¹ ≈ 56 g.mol⁻¹

4 0
3 years ago
Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.2
Rina8888 [55]

Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.

Answer 1) σ Bond a : C has SP^{2}  O has SP^{2} .

Explanation : The orbitals of oxygen and carbon which are involved in SP^{2} hybridization to form sigma bonds. This is observed at 'a' position in the citric acid molecule.

Answer 2) π Bond a: C has π orbitals and  O also has π  orbitals.

Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.

Answer 3) Bond b:  O SP^{3}  H has only S orbital involved in bonding.

Explanation : The bonding at 'b' position involves oxygen SP^{3} hydrogen atoms in it. It has SP^{3} hybridized orbitals and S orbital of hydrogen involved in the bonding.

Answer 4) Bond c:   C is SP^{3}  O is also SP^{3}

Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of SP^{3} hybridized bonds.

Answer 5) Bond d:   C atom has SP^{3}  C atom has SP^{3}

Explanation : At the position of 'd' the bonding between two carbon atoms is found to be SP^{3}. Therefore, the orbitals that undergo SP^{3} hybridization are SP^{3}.

Answer 6) Bond e : C1 containing O SP^{2}    

C2 is SP^{3}

Explanation : The carbon atom which contains oxygen along with a double bond has SP^{2} hybridized orbitals involved in the bonding process; whereas the carbon at C2 has SP^{3} hybridized orbitals involved during the bonding. This is for the 'e' position.

7 0
4 years ago
Explain why electrons orbit around the nucleus
Sonja [21]
In quantum mechanics, an atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus.
6 0
3 years ago
Write a balanced chemical equation for each of the following
stellarik [79]
1.) MgO + Fe --> FeO + Mg
2.) H + I --> HI
3.) Na + I --> NaI
4.) NaO + H2O --> NaOH + H
4 0
4 years ago
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