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3 years ago

As you move from left to right across the periodic table elements ?

Chemistry

1 answer:

Sidana [21]3 years ago

6 0

Explanation:

Quite a number of properties varies across a period. Some remains constant whereas others decreases.

As one moves from left to right;

The energy level remains the same.
The ionization energy increases progressively as a result of increasing nuclear charge.
Electron affinity increases from left to right.
Electronegativity increases.
Electropositivity decreases.

learn more:

Periodic table brainly.com/question/2014634

#learnwithBrainly

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C₂H₄ + 3O₂ → 2CO₂ + 2H₂O If you start with 45.9 g of C₂H₄, and excess O₂, what mass of CO₂ will be produced?

hoa [83]

The mass of CO2 that would be produced is 144.03 g

From the equation of the reaction, the mole ratio of C2H4 input to CO2 produced is 1:2.

Recall that: mole = mass/molar mass

mole of 45.9 g C2H4 = 45.9/28.05

= 1.6364 moles

Thus: equivalent moles of CO2 = 1.6364 x 2

= 3.2727 moles

Mass of 3.2727 moles of CO2 = moles x molar mass

= 3.2727 x 44.01

= 144.03 g

More on stoichiometric calculations can be found here: brainly.com/question/8062886?referrer=searchResults

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3 years ago

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3 years ago

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Valentin [98]

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3 years ago

Why is plastic not classified as a composite?

pantera1 [17]

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3 years ago

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allochka39001 [22]

Answer:

$\large \boxed{\text{0.1767 mol/L}}$$

Explanation:

(a) Balanced equation

3NaOH + H₃PO₄ ⟶ Na₃PO₄ + 3H₂O

(b) Moles of NaOH

$\text{Moles of NaOH} = \text{34.52 mL NaOH} \times \dfrac{\text{0.3840 mmol NaOH}}{\text{1 mL NaOH}} = \text{13.26 mmol NaOH}$

(c) Moles of H₃PO₄

The molar ratio is 1 mol H₃PO₄:3 mol NaOH.

$\text{Moles of H$_{3}$PO}_{4} = \text{13.26 mmol NaOH} \times\dfrac{\text{ 1 mmol H$_{3}$PO}_{4}}{\text{3 mmol NaOH}}\\\\= \text{4.419 mmol H$_{3}$PO}_{4}$

(d) Molar concentration of H₃PO₄

$c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V} = \dfrac{\text{4.419 mmol}}{\text{25.00 mL}} = \text{0.1767 mol $\cdot$ L$^{-1}$}\\\\\text{The molar concentration of the NaOH is $\large \boxed{\textbf{0.1767 mol/L}}$}$

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3 years ago