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frez [133]
3 years ago
8

How many double bonds are in CO2? O 1 o 2 O 3

Chemistry
1 answer:
telo118 [61]3 years ago
7 0

Explanation:

In CO₂ there's 2 double bonds.

Structure:

O=C=O (linear) (180°)

Therefore,

Option C is correct✔

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True or false: This balanced equation shows the combustion of ethane: 2C2H6 + 7O2 → 4CO2 + 6H2O
MakcuM [25]

2C2H6 + 7O2 → 4CO2 + 6H2O

C2H6 is ethane,

Ethane reacts with oxygen, so it is the combustion.

The equation is balanced.

Answer is True.

5 0
4 years ago
A chemist prepares repares a solution of magnesium chloride (MgCl2) by measuring out 48. mg of MgCl, into a 300 ml. volumetric f
Elodia [21]

Answer:

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8 0
3 years ago
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Among the following options, a valid Lewis structure of __________ cannot be drawn without violating the octet rule.
Murrr4er [49]

Answer:

d. IF3

Explanation:

The Octet rule posits that atoms gain, atom lose, or share electrons in order to have a full valence shell of 8 electrons. This statement occurs when atoms also combine to form molecules until they attain or share eight valence electrons either by losing or gaining eletrons.

From the given options, a valid Lewis structure that cannot be drawn without violating the octet rule is IF3

8 0
3 years ago
If 175 g of phosphoric acid reacts with 150.0 g of sodium hydroxide, what is the limiting reactant? How many grams of sodium pho
Evgen [1.6K]

Answer:

NaOH is the limiting reactant.

204.9 g of sodium phosphate are formed.

51.94 g of excess reactant will remain.

Explanation:

The reaction that takes place is:

  • H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O

First we <u>convert the mass of both reactants to moles</u>, using their <em>respective molar masses</em>:

  • H₃PO₄ ⇒ 175 g ÷ 98 g/mol = 1.78 mol
  • NaOH ⇒ 150 g ÷ 40 g/mol = 3.75 mol

1.78 moles of H₃PO₄ would react completely with (1.78 * 3) 5.34 moles of NaOH. There are not as many NaOH moles so NaOH is the limiting reactant.

--

We <u>calculate the produced moles of Na₃PO₄</u> using the <em>limiting reactant</em>:

  • 3.75 mol NaOH * \frac{1molNa_3PO_4}{3molNaOH} = 1.25 mol Na₃PO₄

Then we <u>convert moles into grams</u>:

  • 1.25 mol Na₃PO₄ * 163.94 g/mol = 204.9 g

--

We calculate how many H₃PO₄ moles would react with 3.75 NaOH moles:

  • 3.75 mol NaOH * \frac{1molH_3PO_4}{3molNaOH} = 1.25 mol H₃PO₄

We substract that amount from the original amount:

  • 1.78 - 1.25 = 0.53 mol H₃PO₄

Finally we <u>convert those remaining moles to grams</u>:

  • 0.53 mol H₃PO₄ * 98 g/mol = 51.94 g
3 0
3 years ago
If 0.278g of argon dissolves in 1.5 l of water at 62 bar, what quantity of argon will dissolve at 78 bar
irinina [24]
When P1/P2 = C1/C2

and C is the molarity which = moles/volume

so, P1/P2 = [(mass1/mw)/volume] / [(mass2/mw)/volume]

P1/P2 = (mass1/mw)/1.5L / (mass2/mw)/1.5L 

so, Mw and 1.5 L will cancel out:

∴P1/P2 = mass1 / mass2

∴ mass 2 = mass1*(P2 / P1)

                = 0.278g * (78 bar / 62 bar)

                = 0.35 g

∴ the quantity of argon that will dissolve at 78 bar = 0.35 g


5 0
3 years ago
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