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masya89 [10]
2 years ago
13

2. How many moles are in 2.8 Liters of CO2 gas?

Chemistry
1 answer:
andrew-mc [135]2 years ago
4 0

Answer:

0.125 moles

Explanation:

2.8 litres is equivalent to 2.8dm³

At STP,

1 mole = 22.4 dm³

x mole = 2.8 dm³

Cross multiply

22.4x = 2.8

Divide both sides by 22.4

x = 2.8/22.4

x = 0.125

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A sample of propane(c3h8)has 3.84x10^24 H atoms.
deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}

<u>= 4.75 × 10²³ molecules of Propane</u>.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

<u>= 14.25 × 10²³ </u>

<u>________________________________________</u>

<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

\mathsf{ \frac{44 }{6.02 \times  {10}^{23} } \times 4.75 \times  {10}^{23}  }

\mathsf{  = \frac{44 }{6.02 \times   \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}}  }

\mathsf{  = \frac{44 }{6.02 } \times 4.75   }

<u>= 34.72 g</u>

8 0
2 years ago
Draw two glycine molecules and show how they can be linked by a condensation reaction
lawyer [7]
NH₂-CH₂-COOH + HNH-CH₂-COOH → NH₂-CH₂-CO-NH-CH₂-COOH + H₂O
                                                                             amide link
7 0
3 years ago
True or false ? <br>an insoluble base will react with an acid. <br>​
Leno4ka [110]

Answer:

true maybe

Explanation:

#CarryOnLearning

7 0
3 years ago
If you have 1.00 mole of F2 at 1.00 atm of pressure and 0°C, what is the volume of F2?
ss7ja [257]

Answer:

V=22.4L

Explanation:

Hello,

In this case, by using the ideal gas equation, er can compute the volume of fluorine gas as shown below:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{1mol*0.082\frac{atm*L}{mol*K}+(0+273)K}{1.00atm}  \\\\V=22.4L

Best regards.

4 0
3 years ago
Which elements did the Curies discover?
Scilla [17]

Answer:

B on edge

Explanation:

6 0
3 years ago
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