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antoniya [11.8K]
3 years ago
14

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

Chemistry
1 answer:
stellarik [79]3 years ago
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
  
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3 years ago
Consider two gases, A and B, are in a container at room temperature. What effect will the following changes have on the rate of
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Answer:

B: increase.

Explanation:

When we are considering two gases A and B in a container at room temperature .

We have to find the change on  rate of reaction when the number of molecules of gases A is doubled

Let [A]=a and [B]=b

A+B\rightarrow product

Rate of reaction

R_1=k[A][B]=kab

We know that concentration is increases with increase in number of moles

When the number of molecules of gases A is doubled then concentration of gases A increases.

Therefore ,[A]=2a

Rate of reaction

R_2=k(2a)(b)=2kab

R_2=2R_1

Hence, the rate of reaction is  2 times the initial rate of reaction.Therefore, the rate of reaction will increase when the number of molecules of gases A is doubled.

Answer: B: increase.

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2 years ago
Cómo se forma el enlace ionico​
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5 0
3 years ago
Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re
borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

where; a = 78.2 ; b = 1 ; c= - 0.0530

= \frac{-b+\sqrt{b^2-4ac} }{2a}   or \frac{-b-\sqrt{b^2-4ac} }{2a}

= \frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}  or \frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}

= 0.0204  or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

     = 0.0530 - 0.0204

     = 0.0326

Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K  (fixed constant temperature )

V (volume) = ???

V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

V = 1.64604

V ≅ 1.65 L

3 0
3 years ago
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