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alexira [117]
3 years ago
13

50 compounds in chemistry​

Chemistry
1 answer:
Lana71 [14]3 years ago
5 0

Answer:

\Large \boxed{\mathrm{view \ explanation}}

Explanation:

Chemical compounds are substances that contain two or more elements that are chemically bonded together.

Here are 50 chemical compounds:

  1. Hydrogen peroxide
  2. Fructose
  3. Sulfur hexafluoride
  4. Glucose
  5. Sulfuric acid
  6. Calcium nitrate
  7. Nitrous oxide
  8. Calcium sulfate
  9. Ammonia
  10. Hydrochloric acid
  11. Potassium nitrate
  12. Potassium nitrite
  13. Sucrose
  14. Methane
  15. Carbon dioxide
  16. Dihydrogen oxide
  17. Putrescine
  18. Chloroform
  19. Ethene
  20. Hydrazine
  21. Sodium bicarbonate
  22. Sodium chloride
  23. Acetate
  24. Magnesium sulfate
  25. Urethane
  26. Butyne
  27. Nicotine
  28. Maltose
  29. Propene
  30. Tartaric acid
  31. Nitrogen dioxide
  32. Butane
  33. Butene
  34. Propane
  35. Pentane
  36. Hexane
  37. Benzene
  38. Isobutane
  39. Ovalene
  40. Coronene
  41. Pyrene
  42. Chrysene
  43. Napthalene
  44. Acetic acid
  45. Barium iodide
  46. Aluminium oxide
  47. Aluminium fluoride
  48. Styrene
  49. Toluene
  50. Vinyl chloride
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What is the mass of 1.7 × 1023 atoms of zinc
fiasKO [112]

We are given with a compound, Zinc (Zn) having a 1.7 x 10 ^23 atoms. We are tasked to solve for it's corresponding mass in g. We need to find first the molecular weight of Zinc, that is

Zn= 65.38 g/mol

Not that 1 mol=6.022x10^{23} atoms, hence,

1.7 x 10 ^23 atoms x 1 mol/6.022x10^{23} atoms x65.38 g/ 1mol

=18.456 g of Zn

 

Therefore, the mass of Zinc 18.456 g

3 0
3 years ago
Consider the reaction. PCl5(g)↽−−⇀PCl3(g)+Cl2(g) K=0.042 The concentrations of the products at equilibrium are [PCl3]=0.18 M and
tatyana61 [14]

<u>Answer:</u> The equilibrium concentration of PCl_5 is 1.285 M.

<u>Explanation:</u>

The chemical equation for the decomposition of phosphorus pentachloride follows:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

The expression for equilibrium constant is given as:

K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}

We are given:

K_c=0.042

[PCl_3]=0.18M

[Cl_2]=0.30M

The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Putting values in above equation, we get:

0.042=\frac{0.18\times 0.30}{[PCl_5]}

[PCl_5]=1.285

Hence, the equilibrium concentration of PCl_5 is 1.285 M.

6 0
3 years ago
Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0
3 years ago
Hurry need answer as soon as possible please . Compare the type of change that occurs when a substance melts and when a substanc
tamaranim1 [39]

Answer:

Explanation:

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On the other hand, burning is a chemical change which produces an entirely new compount(in most cases, ash) which does not have the same properties as the original object

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