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3 years ago

PLEASE HELP <3

1 answer:

8 0

In order for carbon to be stable and have 8 electrons, it must make 4 total covalent bonds.

In prefer for oxygen to be stable and have 8 electrons, it must make 2 covalent bonds.

So, we can deduce that CO2 looks like this:

O=C=O

This molecule has two double bonds.

Pssst...Can I get a brainliest?

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Aneli [31]

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5 0

3 years ago

Read 2 more answers

Assuming that a tank of gasoline contains 80 liters and that its density is 0.77 kg/liter, determine how many kg of co2 are prod

sleet_krkn [62]

Answer: -

If a tank of gasoline contains 80 liters and that its density is 0.77 kg/liter, 0.26 kg of CO₂ are produced for each tank of gasoline burned.

Explanation: -

Density of the gasoline = 0.77 kg / liter

Volume of the tank containing the gasoline = 80 liter.

Mass of gasoline produced from each tank

= Volume of the tank containing the gasoline x Density of the gasoline

= $\frac{0.77 kg}{1 liter}$ x 80 liter

= 61.6 kg

Chemical formula of gasoline = C₈H₁₈

Molar mass of gasoline C₈H₁₈ = 12 x 8 + 1 x 18 = 114 g/ mol

Number of moles of C₈H₁₈ = $\frac{61.6 g}{114 g}$ x 1 mol

= 0.54 mol of C₈H₁₈

The chemical equation for the burning of gasoline is

2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

From the balanced equation we see

2 mol of C₈H₁₈ gives 16 mol of CO₂

0.54 mol of C₈H₁₈ gives $\frac{16 mol CO2 x 0.54 mol C8H18}{2 mol C8H18}$ mol of CO₂

= 4.32 mol of CO₂

Molar mass of CO₂ = 12 x 1 + 16 x 3 =60 g / mol

Mass of CO₂ = Molar mass of CO₂ x Number of moles of CO₂

= $\frac{60g x 4.32 mol}{1 mol}$

= 259.2 g

= $\frac{259.2}{1000}$

= 0.259 Kg

= 0.26 kg rounded off to 2 significant figures.

Thus if a tank of gasoline contains 80 liters and that its density is 0.77 kg/liter, 0.26 kg of CO₂ are produced for each tank of gasoline burned.

4 0

3 years ago

A phosphate buffer solution (25.00 mL sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The compone

AysviL [449]

Answer:

0,07448M of phosphate buffer

Explanation:

sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:

Na₂HP + HCl ⇄ NaH₂P + NaCl (1)

NaH₂P + HCl ⇄ H₃P + NaCl (2)

The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:

0,01862L of HCl× $\frac{0,1000mol}{L}$ = 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.