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attashe74 [19]
3 years ago
7

how many moles of cesium carbonate will be produced when 5.34 moles of cesium reacts with iron iii carbonate

Chemistry
1 answer:
geniusboy [140]3 years ago
8 0

Answer:

                      2.67 moles of Cs₂CO₃

Explanation:

                     The balance chemical equation for given single replacement reaction is;

                            6 Cs + Fe₂(CO₃)₃ → 3 Cs₂CO₃ + 2 Fe

According to equation,

                       6 moles of Cs produced  =  3 moles of Cs₂CO₃

So,

                       5.34 moles of Cs will produce  =  X moles of Cs

Solving for X,

                       X  =  5.34 mol × 3 mol / 6 mol

                      X  =  2.67 moles of Cs₂CO₃

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7 0
3 years ago
Can someone help me understand what 1b is asking and how to answer it?
icang [17]
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5 0
3 years ago
1- Alum used in cooking is potassium aluminum sulfate hydrate, KAl(SO4)2. XH2O. To find the value of X, you can heat the sample
Burka [1]

Answer:

THE VALUE OF X IS 7 AND THE FORMULA OF THE HYDRATED SALT IS KAl(SO4)2.7H20

Explanation:

1. write out the varibales given in thequestion:

Mass of the hydrated salt = 4.74 g

Mass of water lost = 2.16 g

Formula of the hydrated salt = KAl(SO4)2. XH20

2. calculate the molar mass of the salt and that of water of crystallization:

Molar mass of anhydrous salt = ( K = 39, Al = 27, S = 32, 0=16)

= ( 39 + 27 + 32*2 + 16 * 8

= (39 + 27 + 64 + 128)

= 258 g/mol

Molar mass of water = 18 g/mol

3. Use this expression to calculate X:

The expression,

XH20 / molar mass of anhydrous salt = Mass of water lost / Mass of hydrated salt.

X = molar mass of anhydrous salt * mass of water lost / mass of anhydrous salt * H20

where XH20 is the molar mass of water of crystallization, is used to calculate the value of X.

4. Solve for X:

So therefore:

X = 258 * 2.16 / 4.74 * 18

X = 557.28/ 85.32

X = 6.53

X is approximately 7.

The value of X is 7 and the formula pf the hydrated salt is KAl(SO4)2.7H20

7 0
3 years ago
How many grams of glucose are needed to prepare 400. g of a 2.00% (m/m) glucose solution g?
Dominik [7]
The grams   of glucose  are  needed  to  prepare  400g  of  a 2.00%(m/m)  glucose  solution  g  is  calculated  as  below

=% m/m =mass  of the solute/mass  of  the  solution  x100

let mass of   solute  be represented  by  y
mass  of solution = 400 g
 % (m/m) = 2% = 2/100

 grams  of  glucose  is  therefore =2/100 =  y/400
by cross  multiplication

100y = 800
divide   both side  by  100

y= 8.0 grams



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