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Nadusha1986 [10]

3 years ago

11

Whats is tge electron configuration of Ag2+ please answer asap..

1 answer:

Triss [41]3 years ago

6 0

Answer:

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10

Explanation:

To write the configuration for the Silver and the Silver ion, first we need to write the electron configuration for just Silver (Ag). We first need to find the number of electrons for the Ag atom (there are 47 electrons) using the Periodic Table. When we write the configuration, we'll put all 47

Either way, the Silver electron configuration will be 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s1

Note that when writing the electron configuration for an atom like Ag and Ag+, the 4d is usually written before the 5s. Both of the configurations have the correct numbers of electrons in each orbital, it is just a matter of how the electronic configuration notation is written (see below for an explanation why).

For the Ag+, called the Silver ion, we remove one electron from 5s2 leaving us with:

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10

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How many moles of electrons are transferred when 2.0 moles of aluminum metal react with excess copper(II) nitrate in aqueous sol

mojhsa [17]

Moles of electrons:

The moles of electrons that are transferred are 12F

A balanced equation:

2 moles of Aluminium metal react with excess copper(II) nitrate.

$2Al + 3Cu{(NO_3)}_2 \rightarrow 2Al{(NO_3)}_3 +3 Cu$

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Moles of Aluminium = 2

As Aluminium goes from 0 to +3 oxidation state

$Al \rightarrow Al^{3+} + 3e^-$

And copper goes from +2 to 0

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Total $(2\times6)$ F mole of electrons

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4 0

1 year ago

The The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k=0.029 W/ m measured temperature differ

vfiekz [6]

Answer:

Heat flux = 13.92 W/m2

Rate of heat transfer throug the 3m x 3m sheet = 125.28 W

The thermal resistance of the 3x3m sheet is 0.0958 K/W

Explanation:

The rate of heat transfer through a 3m x 3m sheet of insulation can be calculated as:

$q=-k*A*\frac{\Delta T}{\Delta X}\\\\q=-0.029\frac{W}{m*K}*(3m*3m)*\frac{12K}{0.025m} =125.28W$

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It can also be calculated as:

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The thermal resistance can be expressed as

$\Delta T=R_t*Q\\R_t=\Delta T/Q=\frac{\Delta X}{k*A}$

For the 3m x 3m sheet, the thermal resistance is

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