Fill in the coefficients that will balance the following reaction:

Chemistry

2 answers:

Tema [17]3 years ago

6 0

Answer:

a0=1

a1=6

a2=2

a3=3

Explanation:

We have to start with the reaction:

$Cr_2(SO_4)_3~+~AgNO_3~->~Cr(NO_3)_3~+~Ag_2SO_4$

We have to start with " $Cr$ ". If we want to balance Cr we have to obtain 2 on both sides. Therefore we have to add a "2" on the right side, so:

$Cr_2(SO_4)_3~+~AgNO_3~->~2Cr(NO_3)_3~+~Ag_2SO_4$

Then we can balance " $NO_3$ " as a whole. We have 6 on the right, we have to have the same amount on the left side, so:

$Cr_2(SO_4)_3~+~6AgNO_3~->~2Cr(NO_3)_3~+~Ag_2SO_4$

Then we can balance " $Ag$ ". We have 6 on the left, so we have to have the same amount on the right if we already have 2 we have to put a · in front (on the right side) to obtain in total 6, so:

$Cr_2(SO_4)_3~+~6AgNO_3~->~2Cr(NO_3)_3~+~3Ag_2SO_4$

Finally, when we add this last number the $SO_4$ we will have 3 on both sides. Therefore the reaction is already balanced.

Kay [80]3 years ago

3 0

Answer

Explanation:

can you please put it in boxes so i can find out the answer

You might be interested in

Which sentences describe decomposers in a food chain?

Elan Coil [88]

Answer:

hey are the final link in the energy flow in a food chain or a food web. They are fungi and animals that feed on dead organic mattet.

Explanaticuz im naruto uzamaki

5 0

2 years ago

: An unknown metal crystallizes in the cubic crystal structure. The metal has a radius 140pm, atomic mass of 135 g/mol, and dens

ki77a [65]

The cubic unit cell this metal crystallize as is BCC structure .

The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell.

The unit cell consists of lattice points that represent the locations of atoms or ions.

The entire structure then consists of this unit cell repeating in three dimensions

$\rm \rho =\dfrac{nM}{N_{0} a^{3}} \\\\\\\\\rm n =\dfrac{\rho N_{0} a^{3}}{M} \\\\\\\\Assuming\; it \; to \; be \; a\; BCC\; structure \\\\\\ r = \dfrac{\sqrt{3} \times a}{4} \\\\\\Therefore\; a = 3.23 \times 10^{-8}\\\\\\\\\rm n =\dfrac{13.3 \times 6.022 \times 10^{23}\times 3.23^{3}\times (10^{-8})^{3}}{135} \\\\$