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mars1129 [50]
3 years ago
13

Given n2 (g) + 3 h2 (g) 2 nh3 (g), which scenario will allow you to eventually reach an equilibrium mixture involving these chem

icals?
Chemistry
1 answer:
mash [69]3 years ago
5 0

<span>The only scenario that will allow you to reach an equilibrium mixture involving these chemicals is to place NH3 into a sealed vessel. This reaction requires pressures between 2100, 3600 psi, and temperatures between 300 and 550 degree Celsius. With this given temperature and pressure, the ammonia naturally decomposes into nitrogen and hydrogen gas at the same rate. When this happen, the concentrations of these chemicals become constant and the system is said to be at equilibrium.</span>

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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
stellarik [79]

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

7 0
4 years ago
How many atoms of oxygen are present in 7.51 grams of<br> glycine with formula C₂H5O2N?
Blizzard [7]

1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.

How to calculate number of atoms?

The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

However, the number of moles of oxygen in glycine can be calculated using the following expression:

Molar mass of C₂H5O2N = 75.07g/mol

Mass of oxygen in glycine = 32g/mol

Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine

Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles

Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms

Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.

Learn more about number of atoms at: brainly.com/question/8834373

#SPJ1

3 0
2 years ago
______________ have properties of both metals and non metals.
stepan [7]

Answer:

gold

Explanation:

4 0
3 years ago
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vovangra [49]
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5 0
3 years ago
Guys, look at the images. I'm so stuck, and I feel as if my brain is about to get ripped out.
Rama09 [41]

Answer:

You not alone lolI'm also tryna figure out the answer

3 0
3 years ago
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