All categories

4 years ago

A hydrate of iron (ii) sulfite is known to contain 44.26% mater. what is the correct name of the hydrate

1 answer:

6 0

The correct name of the hydrate iron (ii) sulfite is iron(ii)sulfite heptahydrate
it is gotten as below

find the % composition of FeSO4 = 100-44.26 =55.74 %

find the mole of FeSO4 and H20
= % composition/molar mass

FeSo4 =55.74/151.9 = 0.367 moles
H2O = 44.26/18 =2.459 moles

find the mole ratio by diving each mole with smallest mole
FeSO4 = 0.367/0.367 = 1
H2O = 2.459/0.367 = 7

therefore the name is iron (ii) sulfite heptahydrate is it has seven water molecules

You might be interested in

How would you dispose the following waste materials

ELEN [110]

To dispose natural waste :

insted of wasting this natural waste u can use them as fertilizer to the plants

To dispose plastic waste : 

recycling , reuse and reduce is the process

hey ya mate !!

hope it helps!! (ʘᴗʘ✿)

8 0

3 years ago

Which equation best represents the balanced net ionic equation for the reaction that occurs when aqueous solutions of lithium ph

forsale [732]

Answer:

2 PO₄³⁻(aq) + 3 Mg²⁺(aq) ⇒ Mg₃(PO₄)₂(s)

Explanation:

Let's consider the molecular equation that occurs when aqueous solutions of lithium phosphate and magnesium nitrate are mixed.

2 Li₃PO₄(aq) + 3 Mg(NO₃)₂(aq) ⇒ 6 LiNO₃(aq) + Mg₃(PO₄)₂(s)

The complete ionic equation includes all the ions and insoluble species.

6 Li⁺(aq) + 2 PO₄³⁻(aq) + 3 Mg²⁺(aq) + 6 NO₃⁻(aq) ⇒ 6 Li⁺(aq) + 6 NO₃⁻(aq) + Mg₃(PO₄)₂(s)

The net ionic equation includes only the ions that participate in the reaction and insoluble species.

2 PO₄³⁻(aq) + 3 Mg²⁺(aq) ⇒ Mg₃(PO₄)₂(s)

8 0

3 years ago

Cuantos gramos de sal pueden disolverse en 50 gramos de agua?

astra-53 [7]

How many grams of salt can be dissolved in 50 grams of water

7 0

4 years ago

which statement regarding the relative energies of monochromatic light with with λ = 800 nm andmonochromatic light with λ = 400

olga2289 [7]

Answer:

The correct answer is option A.

Explanation:

Energy of the photon of an electromagnetic radiation is give by Planck's equation:

$E=\frac{hc}{\lambda }$

Where:

h = Planck's constant

c = speed of light

$\lambda$ = Wavelength of the photon's radiation

Energy of photon of 800 nm monochromatic light

$\lambda = 800 nm=400\times 10^{-9} m$

$E=\frac{hc}{800 \times 10^{-9} m}$

Energy per mole of photons:

$E\times N_A=\frac{hc}{800 \times 10^{-9} m}\times 6.022\times 10^{23} mol^{-1}$ ..[1]

Energy of photon of 400 nm monochromatic light

$\lambda '= 400 nm=400\times 10^{-9} m$

Energy per mole of photons:

$E'\times N_A=\frac{hc}{400 \times 10^{-9} m}\times 6.022\times 10^{23} mol^{-1}$ ..[2]

[1] ÷ [2]

$\frac{E\times N_A}{E'\times N_A}=\frac{\frac{hc\trimes 6.022\times 10^{23} mol^{-1}}{800 \times 10^{-9} m}}{\frac{hc\times 6.022\times 10^{23} mol^{-1}}{400 \times 10^{-9} m}}$