$10800 C$ of electricity through the electrolyte deposited $2.977 g$ of metal with atmic mass $106.4 g m o l^{- 1}$ . The charge on the metal cation is:

+ 4

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Solution

The correct option is A $+ 4$
$10800 C = 2.977 g$

$96500 C = ?$
Hence,

$\frac{96500 \times 2.977}{10800}$
$= 26.600 g / m o l$
$t h u s, \frac{106.4 g / m o l}{26.600 g / m o l}$
$= + 4$ is the answer.

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Similar questions

10800 C

of electricity through the electrolyte deposited

2.977 g

of metal with atomic mass

106.4 g m o l^{- 1}

. The charge on the metal cation is

10800 C

of electricity on passing through the electrolyte solution deposited

2.977

of metal with atomic mass

{106.4 g m o l}^{- 1}

the charge on the metal cation is:

10800C ofelectricity through the electrolyte deposited 2. 977 gram of metal with atomic mass of 106.4 gram per mole find the charge on the metal cations

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10800 C of electricity through the electrolyte deposited 2.977 g of metal with atmic mass 106.4 g mol−1. The charge on the metal cation is:

The correct option is A +410800C=2.977g96500C=?Hence, 96500×2.97710800=26.600g/molthus,106.4g/mol26.600g/mol=+4 is the answer.

$10800 C$ of electricity through the electrolyte deposited $2.977 g$ of metal with atmic mass $106.4 g m o l^{- 1}$ . The charge on the metal cation is:

The correct option is A $+ 4$
$10800 C = 2.977 g$

$96500 C = ?$
Hence,

$\frac{96500 \times 2.977}{10800}$
$= 26.600 g / m o l$
$t h u s, \frac{106.4 g / m o l}{26.600 g / m o l}$
$= + 4$ is the answer.