All categories

3 years ago

Iron (ii) hydroxide + sodium nitrate. Name the precipitate formed in reaction

Chemistry

1 answer:

olya-2409 [2.1K]3 years ago

4 0

Green colored precipitate of iron (||) hydroxide

You might be interested in

Compared with the fibers of cotton plants growing today, what is the relative ratio radioactivity in the old material vs the rel

TiliK225 [7]

Answer:

0.56

Explanation:

From the formula;

0.693/t1/2 = 2.303/t log (Ao/At)

t1/2 = half life of the C-14 = 5730 y

t = time elapsed = 4800 y

At = Activity of C-14 at time t

Ao= Activity of a living C-14 sample

0.693/5730 = 2.303/4800 log (Ao/At)

1.2 * 10^-4 = 4.8 * 10^-4 log (Ao/At)

log (Ao/At) = 1.2 * 10^-4/4.8 * 10^-4

log (Ao/At) = 0.25

Ao/At = Antilog (0.25)

Ao/At = 1.778

Hence;

At/Ao = (1.778)^-1

At/Ao = 0.56

7 0

3 years ago

QUICK!!! Justin is walking around an area that has few trees and many tall grasses.

romanna [79]

Justin is most likely walking in a grassland biome .

6 0

2 years ago

The leading cause of air pollution is _____. automobiles solar power furnaces cigarette smoke

Naddika [18.5K]

Automobiles have they highest cost for of pollution in the air

8 0

3 years ago

Read 2 more answers

Lipids (also known as fats) have two types of monomers called

dalvyx [7]

Fats are large molecules made of two types of molecules, glycerol and some type of fatty acid.

4 0

2 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

2 years ago