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2 years ago

I need an answer now please asap and you will me marked brainiest please it's missing I need it now.

Chemistry

2 answers:

Lesechka [4]2 years ago

7 0

put 2 in front of hydrogen and 1 in front of oxygen and 2 in front of water; to<em> balance the equation</em>.

Phoenix [80]2 years ago

3 0

Answer:

Just put 1 in the boxes

Explanation:

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A Cu2+ solution is prepared by dissolving a 0.4749 g piece of copper wire in acid. The solution is then passed through a Walden

Luda [366]

Answer:

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

Concentration of $MnO_4^-$ = 0.03721 M

Explanation:

The reduction for $Cr_2O_7^{2-}$ is;

$Cr_2O_7^{2-} + 14 H ^+ _{(aq)} + 6 e^- -----> 2 Cr^{3+} _{(aq)}+7H_2O _{(l)}$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

6 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $Cr_2O_7^{2-}$ reacted = $\frac{0.00747}{6}$

number of moles of $Cr_2O_7^{2-}$ reacted = 0.001245 mole

Concentration of $Cr_2O_7^{2-}$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $Cr_2O_7^{2-}$ = $\frac{0.001245}{40.15*10^{-3}}$

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

The reduction for $MnO_4^-$ is;

$MnO_4^- + 8H^+ + 5 e^- -----> Mn^{2+} + 4H_2O$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

5 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $MnO_4^-$ reacted = $\frac{0.00747}{5}$

number of moles of $MnO_4^-$ reacted = 0.001494 mole

Concentration of $MnO_4^-$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $MnO_4^-$ = $\frac{0.001494 }{40.15*10^{-3}}$

Concentration of $MnO_4^-$ = 0.03721 M

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4 years ago

Write net ionic equations for the reaction, if any, that occurs when aqueous solutions of the following are mixed. ammonium sulf

aleksandr82 [10.1K]

Answer:

See explanation for details

Explanation:

ammonium sulfate and barium nitrate Answer ‍

Ba2+(aq) + SO4^2-(aq) ----------> BaSO4(s)

lead(II) nitrate and sodium chloride Answer ‍

Pb2+(aq) + 2Cl-(aq) -----------> PbCl2(s)

copper(II) chloride and sodium hydroxide Answer ‍ ‍ ‍ ‍ ‍

Cu2+(aq) + 2OH-(aq) -------------> Cu(OH)2(s)

There can only be a net ionic reaction when a precipitate is formed as the two solutions are mixed together.

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When alpha particles are used to bombard gold foil, most of the alpha particles pass through undeflected. this result indicates

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When alpha particles are used to bombard gold foil, most of the alpha particles pass through undeflected. This result of the experiment indicates that most of the volume of a gold atom consists of unoccupied spaces. This was an experiment done by Ernest Rutherford which allowed him to verify the presence of the a small region positive charge in an atom which is called as the nucleus of the atom. Most of the alpha particles pass through the foil while a certain amount were deflected. Those that pass through confirmed that there is a large area that is unoccupied in an atom.

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