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Sliva [168]
2 years ago
10

The mass lost by the positive electron should equal the mass gained by the negative electrode suggest two reasons why the result

s were not expected.
Chemistry
1 answer:
vesna_86 [32]2 years ago
3 0

Answer:

The gain in mass by the negative electrode is the same as the loss in mass by the positive electrode. So the copper deposited on the negative electrode must be the same copper ions that are lost from the positive electrode.

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A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molec
balu736 [363]

Answer:

E.F= OH

M.F=O_{2} H_{2}

Explanation:

Empirical Formula

Step 1: Calculate mols of each element

0.44gH(1 mol H/ 1.008g H)= 0.4365 mol H

*note leave extra sig figs for calculations

6.92gO(1 mol O/ 16 g O)=0.4325 mol O

Step 2: Identify which is the smallest mol

*in this case 0.4365 mol H> 0.4325 mol O so we will use 0.4325 mol O

Step 3: Divide above calculations by the smallest mol

0.4325 mol O/0.4325= 1

0.4365 mol H/0.4325= 1.009 *rounds to 1

Step 4: use calculations as subscripts

Oxygen = 1 so the subscript will 1 (O)

Hydrogen = 1 so the subscript will 1 (H)

making E.F= OH

Molecular Formula:

Step 1: identify  molecular mass and mass from the E.F

the molecular mass is given 34.00g/mol

the mass of the E.F is

(oxygen mass from periodic table)+(hydrogen mass from periodic table)

16+1.008= 17.008

Step 2:Divide the molecular mass by the mass given by the emipirical formula.

\frac{34.00}{17.008}= 1.999 round to 2

Step 3:Multiply the empirical formula (the subscripts) by this number to get the molecular formula.  ANSWER: M.F=2(OH)- O_{2} H_{2}

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Answer:

n don't hv symmetry lines.

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3 years ago
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