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Firdavs [7]
3 years ago
14

How many molecules are in 100 g of C6H120,?*​

Chemistry
1 answer:
Grace [21]3 years ago
6 0

Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

Number of molecules = ?

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 100 g/ 180.16 g/mol

Number of moles = 0.56 mol

Number of molecules:

1 mole contain 6.022 × 10²³ molecules

0.56 mol × 6.022 × 10²³ molecules /1 mol

3.37 × 10²³ molecules

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wolverine [178]

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3 years ago
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How many atoms of hydrogen are in 1 mole of Dimethylnitrosamine (CH3)2N20?
Alex Ar [27]

there are 6 total Hydrogen atoms in Dimethylnitrosamine

4 0
3 years ago
A mixture of He, Ne, and N2 gases has a pressure of 1.943 atm. If the pressures of He and Ne are 0.137 atm and 0.566 atm, respec
Ad libitum [116K]

Answer:

The parcial pressure of N₂ in the mixture is 1.24 atm.

Explanation:

The pressure exerted by a particular gas in a mixture is known as its partial pressure. So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

PT = PA + PB

This relationship is due to the assumption that there are no attractive forces between the gases.

In this case:

PT=PHe + PNe + PN₂

You know:

  • PT= 1.943 atm
  • PHe= 0.137 atm
  • PNe= 0.566 atm
  • PN₂= ?

Replacing:

1.943 atm= 0.137 atm + 0.566 atm + PN₂

Solving:

1.943 atm= 0.703 atm + PN₂

1.943 atm - 0.703 atm= PN₂

1.24 atm= PN₂

<u><em>The parcial pressure of N₂ in the mixture is 1.24 atm.</em></u>

4 0
2 years ago
If you combine 290.0 mL 290.0 mL of water at 25.00 ∘ C 25.00 ∘C and 140.0 mL 140.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t
Alexxandr [17]

Answer:

The final temperature is 47.79 °C

Explanation:

Step 1: Data given

Sample 1 has a volume of 290.0 mL

Temperature of sample 1 = 25.00 °C

Sample 2 has a volume of 140.00 mL

Temperature of sample 2 = 95.00 °C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qlost = -Qgained

Q = m*c* ΔT

Q(sample1) = -Q(sample2)

m(sample1) * c(sample1) * ΔT(sample1) = -m(sample2)*c(sample2) *ΔT(sample2)

⇒with m(sample1) = the mass of sample 1 = 290.0 mL * 1g/mL = 290 grams

⇒with c(sample 1) = the specific heat of water = c(sample 2)

⇒with ΔT(sample 1) = the change of temperature = T2 - T1 = T2 - 25.00 °C

⇒with m(sample2) = the mass of sample 2 = 140.0 mL * 1g/mL = 140 grams

⇒with c(sample2) = the specific heat of water = c(sample1)

⇒with ΔT(sample2) = T2 -T1 = T2 - 95.00°C

m(sample1) *  ΔT(sample1) = -m(sample2)*ΔT(sample2)

290 *(T2-25.0) = -140 *(T2 - 95.0)

290 T2 - 7250 = -140 T2 + 13300

430 T2 = 20550

T2 = 47.79 °C

The final temperature is 47.79 °C

4 0
3 years ago
Complete the following table based on known properties of ionic and covalent compounds.
TiliK225 [7]

Answer:

structure of KF - ionic

Explanation:

6 0
2 years ago
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