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Verizon [17]
3 years ago
14

In an experiment conducted between oxygen and chlorine gas it was found that oxygen gas spread much faster than chlorine. What i

s the explanation for this observation?
Chemistry
1 answer:
soldier1979 [14.2K]3 years ago
4 0

Answer:

Because oxygen is lighter

Explanation:

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Refraction is the bending of waves which are due
Shkiper50 [21]

Answer:

7777

Explanation:

7 0
3 years ago
Formula of a copper (II)sulfate hydrate lab
s344n2d4d5 [400]

Answer:

Weigh the empty crucible, and then weigh into it between 2 g and 3 g of hydrated copper(II) sulphate. Record all weighings accurate to the nearest 0.01 g.

Support the crucible securely in the pipe-clay triangle on the tripod over the Bunsen burner.

Heat the crucible and contents, gently at first, over a medium Bunsen flame, so that the water of crystallisation is driven off steadily. The blue colour of the hydrated compound should gradually fade to the greyish-white of anhydrous copper(II) sulfate. Avoid over-heating, which may cause further decomposition, and stop heating immediately if the colour starts to blacken. If over-heated, toxic or corrosive fumes may be evolved. A total heating time of about 10 minutes should be enough.

Allow the crucible and contents to cool. The tongs may be used to move the hot crucible from the hot pipe-clay triangle onto the heat resistant mat where it should cool more rapidly.

Re-weigh the crucible and contents once cold.

Calculation:

Calculate the molar masses of H2O and CuSO4 (Relative atomic masses: H=1, O=16, S=32, Cu=64)

Calculate the mass of water driven off, and the mass of anhydrous copper(II) sulfate formed in your experiment

Calculate the number of moles of anhydrous copper(II) sulfate formed

Calculate the number of moles of water driven off

Calculate how many moles of water would have been driven off if 1 mole of anhydrous copper(II) sulfate had been formed

Write down the formula for hydrated copper(II) sulfate.

#*#*SHOW FULLSCREEN*#*#

Explanation:

7 0
2 years ago
Gold has a molar (atomic) mass of 197 g/mol. consider a 2.47 g sample of pure gold vapor. (a) calculate the number of moles of g
adell [148]
N = given mass/ molar mass.
n = number of moles
given mass = 2.47 g
molar mass = 197 g/mol

n = 2.47 / 197 
n = 0.01253 moles.
I'm sure you wanted to ask more than this. Just put some comments in. I can do the same.
3 0
3 years ago
Zn + 2 HCl --> ZnCl2 + H2 How many moles of hydrochloric acid (HCl) are used (assume a complete reaction) if 8.3 moles of zin
uysha [10]

Answer:

E. 16.6 mol HCl

Explanation:

The equation for the reaction is;

Zn + 2 HCl --> ZnCl2 + H2

From the reaction 2 moles of HCl produces 1 mole of ZnCl2

Therefore; 8.3 moles of ZnCl2 will be produced by;

  = 8.3 moles ×2

  = 16.6 Moles of HCl

  Therefore;  E.  16.6 mol HCl

4 0
3 years ago
If you burn 55.6 g of hydrogen and produce 497 g of water, how much oxygen reacted?
tester [92]

Answer:

441.28 g Oxygen

Explanation:

  • The combustion of hydrogen gives water as the product.
  • The equation for the reaction is;

2H₂(g) + O₂(g) → 2H₂O(l)

Mass of hydrogen = 55.6 g

Number of moles of hydrogen

Moles = Mass/Molar mass

          = 55.6 g ÷ 2.016 g/mol

          = 27.8 moles

The mole ratio of Hydrogen to Oxygen is 2:1

Therefore;

Number of moles of oxygen = 27.5794 moles ÷ 2

                                               = 13.790 moles

Mass of oxygen gas will therefore be;

Mass = Number of moles × Molar mass

Molar mass of oxygen gas is 32 g/mol

Mass = 13.790 moles × 32 g/mol

<h3>          = 441.28 g</h3><h3>Alternatively:</h3>

Mass of hydrogen + mass of oxygen = Mass of water

Therefore;

Mass of oxygen = Mass of water - mass of hydrogen

                          = 497 g - 55.6 g

<h3>                           = 441.4 g </h3>
6 0
3 years ago
Read 2 more answers
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