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labwork [276]
3 years ago
13

Please help me with this problem!

Chemistry
1 answer:
olganol [36]3 years ago
3 0

Answer: an ex ray and a fire

Explanation:

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For the equilibrium reaction 2so2(g) + o2(g) 2so3(g), δhºrxn = –198 kj/mol. which one of these factors would cause the equilibri
Airida [17]
Let's rewrite the reaction for clarity:

 2 SO₂(g) + O₂(g) ⇆ 2 SO₃(g)            δhºrxn = –198 kj/mol

The equilibrium constant of a reaction is the ratio of the concentration its products to its reactants which are raised to their respective stoichiometric coefficients. For this reaction, the K would be

K = [SO₃]²/[SO₂]²[O₂]

To get a larger K, the products must be greater than the reactants. This means that the forward reaction must be favored to yield more of the product SO₃. There are different ways to do this: by manipulating the pressure, concentration or temperature.

For the concentration, you should add more amounts of the reactants. For the pressure, we should increase it. This is because the product side has only 2 moles of gas compared to 3 moles of gas in the reactants. So, it wall have more room for the product even at a higher pressure. Lastly, since the reaction is exothermic manifested by the negative sign of δhºrxn , the reaction would favor the forward reaction at high temperatures.
5 0
3 years ago
Sugar C12 H22 011 can be classified as a element, heterogeneous mixure, solution, compound​
Fantom [35]

Answer:

The answer is a compound.

it is a compound as it contains different elements

Explanation:

5 0
3 years ago
Gold has an atomic weight of 196.97 g/mol, and a density of 19.3 g/cm3. Approximately how many atoms are in a spherical gold nan
ale4655 [162]

<u>Answer:</u> The number of atoms in spherical gold nano particle are 26.918\times 10^4

<u>Explanation:</u>

  • To calculate the volume of sphere, we use the equation:

V=\frac{4}{3}\pi r^3

where,

r = radius of the sphere = 10nm=10^{-6}cm    (Conversion factor:  1nm=10^{-7}cm )

Putting values in above equation, we get:

V=\frac{4}{3}\times 3.14\times (10^{-6})^3\\\\V=4.19\times 10^{-18}cm^3

  • To calculate mass of of the substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Volume of gold = 4.19\times 10^{-18}cm^3

Density of gold = 19.3g/cm^3

Putting values in above equation, we get:

19.3g/cm^3=\frac{\text{Mass of gold}}{4.19\times 10^{-18}}\\\\\text{Mass of gold}=8.0867\times 10^{-17}g

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of gold = 8.8067\times 10^{-17}g

Molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

\text{Moles of gold}=\frac{8.8067\times 10^{-17}g}{196.97g/mol}=4.47\times 10^{-9}moles

  • According to mole concept:

1 mole of an element contains 6.022\time 10^{23}  number of atom

So, 4.47\times 10^{-19} moles of gold will contain = 4.47\times 10^{-19}\times 6.022\times 10^{23}=26.918\times 10^4 number of atoms.

Hence, the number of atoms in spherical gold nano particle are 26.918\times 10^4

8 0
3 years ago
1. Based on the observed performance of the air bag models and the amounts of sodium bicarbonate and acetic acid (vinegar) neede
Ira Lisetskai [31]

Sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags since the require more mass and produce less gas.

<h3>Which is the better chemical for an airbag?</h3>

The chemical equation for the production of nitrogen gas from sodium azide is given below:

  • NaN₃ → 2 Na + 3 N₂

1 mole or 66 go of sodium azide produces 3 moles or 67.2 L of nitrogen gas.

The equation for the production of carbon dioxide from sodium bicarbonate and acetic acid is given below:

  • Na₂CO₃ + CH₃COOH → CH₃COONa + CO₂ + H₂O

1 mole, 106 g of Na₂CO₃ and 1 mole, 82 g of CH₃COOH are required to produce 1 mole or 22.4 L of CO₂.

The mass of sodium azide required is less than that of sodium bicarbonate and acetic acid required. Also, sodium azide produces a greater volume of gas. Therefore, sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags.

In conclusion, sodium azide is a better choice in airbags.

Learn more about airbags at: brainly.com/question/14954949

#SPJ1

7 0
2 years ago
Which type of solution below has a cloudy appearance? (Choose all that apply)
Gelneren [198K]
B. Suspensions is the answer
8 0
3 years ago
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