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3 years ago

A molecule of butane and a molecule of 2-butene both have the same total number of

1 answer:

8 0

The correct answer is option 1. Butane and 2-butene have the same total number of carbon atoms. They both have four carbon atoms. They differ in there structure since the latter has double bonds on it. As a result of the different structure, they also have different properties.

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What is the molality of a solution of water and kcl if the freezing point of the solution is –3mc030-1.jpgc?

Natasha_Volkova [10]

We will use the expression for freezing point depression ∆Tf
∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is
∆Tf = 0C - (-3°C) = 3°C
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and one Cl- ion:
KCl → K+ + Cl-
Therefore, the molality m of the solution can be calculated as
3 = 2 * 1.86 * m
m = 3 / (2 * 1.86)
m = 0.80 molal

4 0

3 years ago

1. How many protons does 14/6 C contain? What element is this?

Harman [31]

Answer:

It has 6 protons and its Carbon 14

Explanation:

3 0

3 years ago

Identify the gas law that applies to the following scenario: If a gas in a closed container is pressurized from 18.0 atm to 14.0

nalin [4]

Answer:

Gay-Lussac's Law

Explanation:

The pressure is directly proportional to the absolute temperature under constant volume. This states the Gay-Lussac's law. The equation is:

P1T2 = P2T1

<em>Where P is pressure and T absolute temperature of 1, initial state and 2, final state of the gas.</em>

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That means the right option is:

- Gay-Lussac's Law

6 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

Identify the geometry of the complex ion if the hybridization is d2sp3. linear square planar octahedral trigonal planar tetrahed

Leona [35]

Octahedral is the geometry of the complex ion which has d2sp3 hybridization.

The octahedral has eight faces hence the prefix octa. Octahedral molecular geometry describe the shape of compound with six atom or groups of atoms or ligands symmetrically arranged around a central atom.The octahedron is one of the platonic solids.

7 0

4 years ago