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Dima020 [189]
3 years ago
12

श्रीरयाफल्सन यसका लागि तियरलप्रतिस्थापन स्तर को के हो​

Chemistry
1 answer:
jeka943 years ago
3 0
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If white light shines on an object and the red, orange, green, blue, indigo, and violet light is absorbed. What color does your
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Hello
it would be yellow dats missing
6 0
4 years ago
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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:

A)

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

B)

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

3 0
4 years ago
I need help please with this chemistry work ​
Marysya12 [62]

Answer:

question 1: 3

question 2: the number of Valence electrons in the atom

hope it helps

8 0
3 years ago
In an ionic bond:
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The answer I'm pretty sure is a because it can't be d because ionic binding the electrons are affected and it can't be c because that's covalent bonding and it can't be b because they don't swap electrons.
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3 years ago
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Question 1:
CaHeK987 [17]
1) Reaction: 3Mg + N₂ → Mg₃N₂.
m(Mg) = 0,225 g
n(Mg) = 0,225 g ÷ 24,3 g/mol = 0,009 mol.
n(Mg) : n(N₂) = 3 : 1
n₁(N₂) = 0,003 mol.
n₂(N₂) = 0,5331 ÷ 28 = 0,019 mol.
n₃(N₂) = 0,019 mol - 0,003 mol = 0,016, m(N₂) = 0,016mol·28g/mol=0,4467g.
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2) Answer is: 6 <span>of fluorine atoms are combined with one uranium atom.
</span>m(U) = 209 g.
m(F) = 100 g.
n(U) = m(U) ÷ M(U)
n(U) = 209 g ÷ 238 g/mol.
n(U) = 0,878 mol.
n(F) = m(F) ÷ M(F)
n(F) = 5,263 mol
n(U) : n(F) = 0,878 mol : 5,263 mol /:0,878.
n(U) : n(F) = 1 : 6.
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8 0
3 years ago
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