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41 . In the attachment
42.
4 Ru(CO3)3 -------------------> 4 Ru2 O3 + 12 CO2
so 4 moles of ruthenium carbonate will produce 12 moles of CO2
so volume of produced CO2 = 12 x22.4 = 268. 8 litres
or weight of produced CO2 = 12x 44 = 528 gm
43.
First let us know the no. of ions in MgCl2.
The molecular mass of MgCl2 is ( 1 * 24 ) + ( 2 * 35.5 ) = 24 + 71 = 95 a.m.u. .
No. of moles of a substance of atomic mass “a” = (No. of grams in sample)/(molecular mass )
No. of moles in MgCl2 = 245 / 95 = 2.5789474… = 2.579 (approx.) moles .
Now, there are ( 1 * Mg ( +2 ) ) + ( 2 * Cl( -1 ) ) = 3 ions total / molecule .
1 mole = ( Avogradro’s number ) molecules . (Avogadro’s Number = 6.022 * 10 ^ 23)
Therefore , there are a total of 3 * 2.579 * ( Avogadro’s Number ) ions in the sample.
Now ,
The NaCl sample must contain 3 * 2.579 * ( Avogadro’s Number ) ions.
Let its mass = x grams.
Now ,
Molec. Mass of NaCl = ( 1 * 23 ) + ( 1 * 35.5 ) = 58.5 a.m.u. .
No. of moles of NaCl = x/58.5 moles
Now , there are ( 1 * Na ( +1 ) ) + ( 1 * Cl( -1 ) ) = 2 ions / molecule.
Therefore there are a total of 2 * x /58.5* ( Avogadro’s Number ) ions in the NaCl sample.
Therefore ,
3 * 2.579 * ( Avogadro’s Number ) = 2 *x/ 58.5 * ( Avogadro’s Number )
Cancelling the ( Avogadro’s Number ) and transposing ,
x = ( 58.5 *3 * 2.579) /2
Therefore , x = 226 grams approximately
Therefore , we need 226 grams of NaCl to have the same no. of ions as 245 grams of MgCl2 .
42.
4 Ru(CO3)3 -------------------> 4 Ru2 O3 + 12 CO2
so 4 moles of ruthenium carbonate will produce 12 moles of CO2
so volume of produced CO2 = 12 x22.4 = 268. 8 litres
or weight of produced CO2 = 12x 44 = 528 gm
43.
First let us know the no. of ions in MgCl2.
The molecular mass of MgCl2 is ( 1 * 24 ) + ( 2 * 35.5 ) = 24 + 71 = 95 a.m.u. .
No. of moles of a substance of atomic mass “a” = (No. of grams in sample)/(molecular mass )
No. of moles in MgCl2 = 245 / 95 = 2.5789474… = 2.579 (approx.) moles .
Now, there are ( 1 * Mg ( +2 ) ) + ( 2 * Cl( -1 ) ) = 3 ions total / molecule .
1 mole = ( Avogradro’s number ) molecules . (Avogadro’s Number = 6.022 * 10 ^ 23)
Therefore , there are a total of 3 * 2.579 * ( Avogadro’s Number ) ions in the sample.
Now ,
The NaCl sample must contain 3 * 2.579 * ( Avogadro’s Number ) ions.
Let its mass = x grams.
Now ,
Molec. Mass of NaCl = ( 1 * 23 ) + ( 1 * 35.5 ) = 58.5 a.m.u. .
No. of moles of NaCl = x/58.5 moles
Now , there are ( 1 * Na ( +1 ) ) + ( 1 * Cl( -1 ) ) = 2 ions / molecule.
Therefore there are a total of 2 * x /58.5* ( Avogadro’s Number ) ions in the NaCl sample.
Therefore ,
3 * 2.579 * ( Avogadro’s Number ) = 2 *x/ 58.5 * ( Avogadro’s Number )
Cancelling the ( Avogadro’s Number ) and transposing ,
x = ( 58.5 *3 * 2.579) /2
Therefore , x = 226 grams approximately
Therefore , we need 226 grams of NaCl to have the same no. of ions as 245 grams of MgCl2 .
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ANSWER
Molar mass of MgCl2 = 24 + 35.5x2 = 95g/mol
Number of moles of MgCl2 in 245 g = 245/95 = 2.579 moles
1 ion of Mg and 2 ions of Cl
Number of ions = 3 x 2.579 x 6.022x10^23
= 4.659 x 10^24
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
1 Na+ ion and 1 Cl- ion
58.5 g NaCl contain = 2 x 6.022 x 10^23 = 1.2044x10^24
What mass of NaCl will contain 4.659 x 10^24 ions?
Mass of NaCl = (4.659 x 10^24)/ (1.2044x10^24) * 58.5
= 226.3 g
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