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erica [24]
3 years ago
5

Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 5.90 moles of magnesium perchlorate, Mg(ClO4)2.

Chemistry
1 answer:
Lady bird [3.3K]3 years ago
4 0

Answer:

5.90, 11.8, 47.2

Explanation:

Let’s remove the parentheses and write the formula as MgCl₂O₈.

We see that 1 mol Mg(ClO₄)₂ contains 1 mol Mg atoms, 2 mol Cl atoms, and 8 mol O atoms.

∴ \text{Moles of Mg atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{1 mol Mg atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{5.90 mol Mg atoms}  

\text{Moles of Cl atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{2 mol Cl atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{11.8 mol Cl atoms}

\text{Moles of O atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{8 mol O atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{47.2 mol O atoms}

∴ Mg, Cl, O = 5.90, 11.8, 47.2

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2 years ago
What is the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol)?
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the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol) is 1.36g

The reaction is 3 MnO2 + 4 Al ------ 2Al2o3+ Mn

3 mole of manganese oxide give 2 moles of aluminum oxide so by the reaction n( MnO2)/3 =n(al203)2

the formula is n= mass/M so, now substituting values

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The air in a 2 L balloon at 0.998 atm and 34.0 °C. What will be its pressure if it is brought to a higher altitude where it now
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Explanation:

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The combined gas equation is,

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P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 2 L

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T_2 = final temperature of gas = 12.0^oC=273+12.0=285.0K

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\frac{0.998\times 2}{307.0K}=\frac{P_2\times 3.5}{285.0K}

P_2=0.529atm

Thus the pressure if it is brought to a higher altitude where it now occupies 3.5 L and is at 12.0 °C is 0.529 atm

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