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Ghella [55]
3 years ago
5

Copper sulphate is used duringlaboratory preparation of hydrogengas. Give reason.​

Chemistry
1 answer:
tester [92]3 years ago
6 0

Answer:Hydrogen gas is prepared in laboratory by the action of dilute HCl upon granulated zinc. Note: This is a redox reaction since Zn is oxidized and H+ ions is reduced here. Procedure: ... Some crystals of copper sulphate are added to increase the rate of reaction.

Explanation:

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How many moles of butane gas, C4H10, react to produce 2.50 moles of H2O? How many moles of oxygen gas reacts? Given the balanced
viva [34]
The moles  of  butane gas  and  oxygen gas reacted  if  2.50  moles  of H2O is produced  is  calculated  as below

the  equation  for  reaction
2C4H10  +13 O2 = 8CO2 +10 H2O

the moles  of  butane (C4H10) reacted calculation

by  use  of mole ratio between  C4H10: H2O  which  is  2  : 10 the  moles of  C4H10=     2.50  x2/10 = 0.5 moles  of  C4H10  reacted

The  moles of  O2  reacted calculation

by  use of mole  ratio  between   O2 : H2O  which  is  13:10 the moles of O2

=   2.50  x 13/10=  3.25  moles of O2  reacted
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3 years ago
Do you think that there is a high correlation between proficiency in mathematics and proficiency in chemistry? Explain why or wh
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7 0
3 years ago
What uses the most oil in the United States
Furkat [3]

Answer:

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5 0
3 years ago
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o-na [289]

Answer:

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8 0
3 years ago
In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,
Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -              -

At t =equilibrium      (x-s)                s           2s          

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

K_{sp}=s\times {2s}^2

K_{sp}=4s^3

Given  s = 7.4×10⁻³ M

So, Ksp is:

K_{sp}=4\times (7.4\times 10^{-3})^3

K_{sp}=4\times (7.4\times 10^{-3})^3

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -            0.20

At t =equilibrium      (x-s')             s'         0.20+2s'         

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2

Solving for s', we get

<u>s' = 4.0×10⁻⁵ M</u>

<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.</u>

3 0
4 years ago
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