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1. How many moles are in 3.63 x 1023 atoms of Mn?

2. Calculate the molar mass if AlBr3

3. Find the mass of 31.6 moles MnCI2

4. How many moles are in 5.6 L of KNO3?

5. How many moles are in 0.50 grams of CH3OH?

6. How many liters are in 8.20 x 10^14 molecules CsBr?



Sagot :

Ardni313idn

1. 1 mol=6.02.10²³ particles

[tex]\tt \dfrac{3.63\times 10^{23}}{6.02\times 10^{23}}=0.602~moles[/tex]

2. Molar mass = Total atomic mass of component

AlBr₃ = Ar Al + 3. Ar Br

AlBr₃ = 27 + 3. 80 = 267 g/mol

3. mass = mol x MW

MW MnCl₂  = 55 + 2.35.5=126 g/mol

mass = 31.6 x 126 =3981.6 g

4. at STP, 1 mol =22.4 L, so for 5.6 L :

[tex]\tt \dfrac{5.6}{22.4}=0.25~moles[/tex]

5. moles = mass : MW

MW CH₃OH = 12.1 + 4. 1 + 16.1 =32 g/mol

[tex]\tt moles=\dfrac{0.5}{32}=0.016[/tex]

6. 1 mol = 22.4 L at STP

1 mol = 6.02.10²³ particles

moles for  8.20 x 10¹⁴ molecules

[tex]\tt \dfrac{8.2\times 10^{14}}{6.02\times 10^{23}}=1.36\times 10^{-9}[/tex]

volume :

[tex]\tt 1.36\times 10^{-19}\times 22.4=3.05\times 10^{-8}~L[/tex]

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