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Contact [7]
3 years ago
7

a sample of cesium metal reacted completely with water, evolving 48.1 ml of dry H2 at 19C and 768 mmHg. what is the equation for

the reaction? what was the mass of cesium in the sample ?
Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
3 0
The reaction  for  equation is    written  as  follows

2Cs (s) +  2H2O(l) =  2CsOH (aq) +  H2  (g)

The  mass  of  cesium  is  calculated  as  follows

Find  the  moles  of  H2  by use  of  ideal  gas  equation  that is  Pv = nRT
where  n  is  the  number of   moles

 n  =  PV/RT
P=  48.1  ml    in  liters  =  48.1  /1000= 0.0481 l
T=  19 +  273.15 =  292.15K
P=  768mm hg
R (gas  Constant)=  62.364 l.
mmhg/k.mol

n= ( 768mmhg  x0.0481 L) /( 62.364 L.mm hg/k.mol  x  292.15k) =  2.028  x10^-3  moles

by   use  of  mole  ratio from  reacting  equation between  Cs to  H2    which  is   2 :1  the  moles  of Cs  is therefore  =  ( 2.028  x10^-3)  x  2 =  4.05  x10^-3 moles

mass  of  cs is therefore  =  moles  x  molar  mass  of  cs(  132.9g/mol)

=( 4.05  x10^-3)mol  x  132.9 g/mol  =  0.539  grams
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6626 g

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Since density = mass/ volume

mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g

Initial temperature of water = 22°C and final temperature of water = 45°C.

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ΔT water = 45 - 22 = 23°C

For iron:

mass = m,  

specific heat capacity for iron  = 0.444 J/g°C

Initial temperature of iron = 1445°C and final temperature of water = 45°C.

ΔT iron = 45 - 1445 = -1400°C

Quantity of heat (Q) to raised the temperature of a body is given as:

Q = mCΔT

The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.

Q water (gain) + Q iron (loss) = 0

Q water = - Q iron

42800 g ×  4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C

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Ionic compound are usually formed when a metal reacts with a non-metal.

The nomenclature of ionic compounds is given by:

1. Positive ion is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

3. In case of transition metals, the oxidation state are written in roman numerals in bracket in-front of positive ions.

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CuF_2 is an ionic compound because copper element is a metal and fluorine element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The oxidation state of copper in CuF_2 is, (+2)

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