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olga55 [171]
3 years ago
15

How many moles are in 2.3 g of phosphorous

Chemistry
1 answer:
Mice21 [21]3 years ago
3 0
You would find the atomic weight of phosphorus and multiply that 2.3 and get 71.23.
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The energy received by Earth from the sun is what type of energy
DedPeter [7]

Answer:

Solar Energy

Explanation:

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3 years ago
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3.
sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

⇒ n = 0.043 mol

∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

⇒ V2 = 2.4314 L

5.  V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)

⇒ V2 = 0.262 L

6.a)  ∴ V1 = 24.6 L

∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))

⇒ n = 0.0993 mol

∴ T2 = 273 K

∴ P2 = 101.3 KPa = 0.9997 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

3 0
3 years ago
In a motion diagram the size of the object must be less than the distance moved t or f
KATRIN_1 [288]
The awnser is false.
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A forest is cut down to make room for a housing development. Which population is most likely to survive? A. raccoons, which can
Basile [38]
A. Raccoons. They will be able to survive in a housing area just as well, or even better then they had in the forrest.

I hope that helped! :D
5 0
3 years ago
Acenapthalene has the empirical formula C6H5. A solution of 0.515 g of acenapthalene in 15.0 g CHCl3 boils at 62.5oC. The normal
german

Answer:

The molecular formula of an ascenapthalene is C_{12}H_{10}

Explanation:

\Delta T_b=K_b\times m

\Delta T_b=K_b\times \frac{\text{Mass of acenapthalene}}{\text{Molar mass of acenapthalene}\times \text{Mass of chloroform in Kg}}

where,

\Delta T_f =Elevation in boiling point = (62.5-61.7)^oC=0.8^oC

Mass of acenapthalene = 0.515 g

Mass of CHCl_3 = 15.0 g = 0.015 kg (1 kg = 1000 g)

K_b = boiling point constant = 3.63 °C/m

m = molality

Now put all the given values in this formula, we get

0.8^0C=3.67 ^oC/m\times \frac{0.515}{\text{Molar mass of acenapthalene}\times 0.015kg}

\text{Molar mass of acenapthalene}=155.7875 g/mol

Let the molecule formula of the Acenapthalene be C_{6n]H_{5n}

6n\times 12 g/mol+5n\times 1 g/mol=155.7875 g/mol

n = 2.0

The molecular formula of an ascenapthalene is C_{12}H_{10}

4 0
3 years ago
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