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guajiro [1.7K]
4 years ago
10

The permanent electric dipole moment of the water molecule 1H2O2 is 6.2 * 10-30 C m. What is the maximum possible torque on a wa

ter molecule in a 5.0 * 108 N/C electric field?
Chemistry
1 answer:
stellarik [79]4 years ago
7 0

Answer:

3.1 x 10⁻²¹ Nm

Explanation:  

When placed in an external electric filed, an electric dipole experiences a torque. and this torque is represented mathematically with the equation:

torque (τ) = dipole moment vector (P) x electric field vector (E)

τ = P. E . sin θ

where θ is the angle between the water molecule and the electric field, which in this case is 90° (because this is where the torque is maximum)

τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . sin90

τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . 1

solve for τ

τ = 3.1 x 10⁻²¹ Nm

the maximum possible torque on the water molecule is therefore 3.1 x 10⁻²¹ Nm

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D 2,2

Explanation:

We can see that there are 2 chlorines on the reactant side so there has to be a 2 on the product side

Now we have Na + Cl2 --> 2NaCl

The problem now is that there are 2 sodiums on the product side so add a 2 to the Na on the reactant side

2Na + Cl2 --> 2NaCl

Now it's balanced!

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3 years ago
Why can scientists ignore the forces of attraction between particles in a gas under ordinary conditions
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3 0
3 years ago
2.
katovenus [111]

Answer:

A. 6atm

Explanation:

Using pressure law equation:

P1/T1 = P2/T2

Where;

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T1 = initial temperature (K)

P2 = final pressure (atm)

T2 = final temperature (K)

According to this question;

P1 = 3 atm

P2 = ?

T1 = 120K

T2 = 240K

Using P1/T1 = P2/T2

3/120 = P2/240

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240 × 3 = P2 × 120

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4 0
3 years ago
The sun emits electromagnetic waves at many frequencies, but mostly in the infrared, visible, and ultraviolet
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The correct answer is radiant, radiant that get transferred into chemical energy
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4 years ago
Read 2 more answers
How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl
VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3}  }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}

3 0
4 years ago
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