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Olin [163]
2 years ago
10

Name the property of salt investigated in the above experiment

Chemistry
1 answer:
Snezhnost [94]2 years ago
7 0

which experiment please involve a clip

You might be interested in
Consider the following equilibrium:
irga5000 [103]

Consider the following equilibrium:

4NH₃ (g) + 3O₂ (g) ⇄ 2N₂ (g) + 6H₂O (g) + 1531 kJ

The given statement is True, because

According to Le Ch's Principle:

Systems that have attained the state of chemical equilibrium will tend to maintain their equilibrium state.

External factors such as the addition of products and reactants result in the disruption of the equilibrium state.

we expect the system to shift to the direction that offsets the change in concentration.

This results in the state of chemical equilibrium to be reestablished.

Hence, The statement is true,

  • The addition of more ammonia (a reactant) would offset the state of equilibrium.
  • To restore chemical equilibrium, the system must consume the excess reactants to form more products.
  • A shift to favor the products side occurs.

To learn more about equilibrium here

brainly.com/question/3920294

#SPJ4

8 0
2 years ago
20. Stoichiometry is based on
Mariana [72]

Answer:

The correct option is (c)

5 0
3 years ago
Read 2 more answers
If .75 moles of ammonia is needed, how many grams of nitrogen will be consumed?
MrMuchimi

We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.

The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.

Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-

N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)

According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, \frac{1 X 0.75}{2} moles = 0.375 moles of nitrogen is required.

Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.

Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.


5 0
3 years ago
To determine the freezing point depression of a LiCl solution, Toni adds 0.411 g of LiCl to the sample test tube along with 19.7
Cerrena [4.2K]

Answer:

LiCl = 0.492 m

Explanation:

Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.

Our solute is lithium chloride, LiCl.

Our solvent is distilled water.

We do not have the mass of water, but we know the volume, so we should apply density to determine mass.

Density = mass / volume

Density . volume = mass

1 g/mL . 19.7 mL = 19.7 g

We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg

Let's determine the moles of LiCl

0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles

Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m

7 0
2 years ago
Marking anybody who got it the brainliest​
bija089 [108]

I can't open pdf.......

7 0
2 years ago
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