Milli is a prefix used in science and engineering to decribe the number as the exponent x10⁻³. In the prefix kilo, the number is at the exponent x10³.

5.12x10⁵ kilometers are:

5.12x10⁵ kilometers * (1000m / 1km) = 5.12x10⁸ meters

5.12x10² meters * (1m / 1000millimeters) = 5.12x10¹¹ millimeters

8 0

3 years ago

For 2,663 kg of a compound with the formula Al(SO), determine the following quantities (4 pts each); a) The number of moles of t

Nastasia [14]

Answer:

a) 35.485 moles of Al(SO)

b) 35.485 moles of S atoms

c) $2.136197(10^{25})$ Al atoms

d) 567.723 g of O

Explanation:

Let's define the following terms :

1 mol = $6.02.(10^{23})$ elemental units

For example :

1 mol of oxygen atoms = $6.02.(10^{23})$ oxygen atoms

Now, our compound has the following formula

Al(SO)

Where Al is aluminium

S is sulfur

And O is oxygen

All the subscripts are 1 so we can say the following :

1 molecule of Al(SO) has 1 atom of Al , 1 atom of S and 1 atom of O

In terms of moles :

1 mol of Al(SO) has 1 mol of Al , 1 mol of S and 1 mol of O

The molar masses of Al, S and O are

$molarmass_{(Al)}=26.982\frac{g}{mol}$

$molarmass_{(S)}=32.065 \frac{g}{mol}$

$molarmass_{(O)}=15.999\frac{g}{mol}$

If we sum all the molar masses = $(26.982+32.065+15.999)\frac{g}{mol}=75.046\frac{g}{mol}$

Finally, 75.046 g of Al(S0) is 1 mol of Al(SO) which contains 26.982 g of Al, 32.065 g of S and 15.999 g of O.

1 mol of Al(SO) contains 1 mol of Al, 1 mol of S and 1 mol of O.

Now we can calculate a),b),c) and d)

For a)

$2.663 kg=2663g$

75.046 g of Al(SO) = 1 mol of Al(SO)

2663 g of Al(SO) = x

$x=\frac{2663}{75.046}mol=35.485 mol$

2.663 kg of Al(SO) contains 35.485 moles of Al(SO)

b) and c) 1 mol of Al(SO) molecules contains 1 mol of S atoms and 1 mol of Al atoms

We have 35.485 moles of Al(SO) molecules so

We have 35.485 moles of S atoms

And 35.485 moles of Al atoms

If 1 mol = $6.02(10^{23})$

35.485 moles of Al have $(35.485)(6.02)(10^{23})=2.136197(10^{25})$ Al atoms

d) 75.046 g of Al(SO) contains 15.999 g of O

2663 g of Al(SO) contains x g of O

$x=\frac{(2663).(15.999)}{75.046} g$

x = 567.723 g of O

6 0

3 years ago

Draw the structure corresponding to the common name α−chloro−β−methylbutyric acid

uysha [10]

Structure of a-chloro-b-methylbutyric acid (2-chloro-3-methylbutyric acid is attached below.

The structure was drawn using following information.

1) First, the parent chain was identified. The parent chain is containing four carbon atoms, also, this chain belongs to carboxylic acid as the compound name ends with Butyric acid i.e. butyr (Butane) -ic acid (carboxylic acid).

2) Secondly, the positions of substituents were assigned by using the rule which sates that the numbering in carboxylic acid must satrt from the carbonyl carbon. Therefore, the carbon which is attached to carbonyl carbon is named as alpha (2nd position) and carbon next to alpha carbon is named as beta carbon (3rd position). Hence, we add chlorine atom on 2nd carbon next to carbonyl group and methyl group at 3rd carbon with respect to carbonyl carbon.

6 0

3 years ago

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