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3 years ago

12

Why do copper ions go to the negative electrode

1 answer:

Snowcat [4.5K]3 years ago

8 0

Its because the copper ions have a positive charge, therefore, they are attracted to a negative electrode.

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What happens along a transform plate boundary?

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4K + O2 + 2K20<br> Which statement is true?

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How GM food are produced?

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Does a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together? Write the n

olganol [36]

Answer : Yes, a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together.

The net ionic equation will be,

$2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

$CaCl_2(aq)+2HgNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+Hg_2Cl_2(s)$

The ionic equation in separated aqueous solution will be,

$Ca^{2+}(aq)+2Cl^{-}(aq)+2Hg^{+}(aq)+2NO_3^{-}(aq)\rightarrow Hg_2Cl_2(s)+Ca^{2+}(aq)+2NO_3^{-}(aq)$

In this equation, $Ca^{2+}\text{ and }NO_3^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)$

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3 years ago

The air in a 2 L balloon at 0.998 atm and 34.0 °C. What will be its pressure if it is brought to a higher altitude where it now

Phantasy [73]

Answer: 0.529 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 0.998 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 2 L

$V_2$ = final volume of gas = 3.5 L

$T_1$ = initial temperature of gas = $34.0^oC=273+34.0=307.0K$

$T_2$ = final temperature of gas = $12.0^oC=273+12.0=285.0K$

Now put all the given values in the above equation, we get:

$\frac{0.998\times 2}{307.0K}=\frac{P_2\times 3.5}{285.0K}$

$P_2=0.529atm$

Thus the pressure if it is brought to a higher altitude where it now occupies 3.5 L and is at 12.0 °C is 0.529 atm

4 0

4 years ago