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Eva8 [605]
3 years ago
15

The survival of aquatic organisms depends on the small amount of O2 that dissolves in H2O . The diagrams above represent possibl

e models to explain this phenomenon. Which diagram provides the better particle representation for the solubility of O2 in H2O , and why?
Chemistry
1 answer:
koban [17]3 years ago
4 0

Answer:

Diagram 1

Explanation:

The solubility of the oxygen gas in water has to do with the interaction of the oxygen with the dipoles in water.

Water is a polar molecule having oxygen as the negative dipole and hydrogen as the positive dipole.

Water can interact with the oxygen atoms in the molecule via intermolecular hydrogen bonds with molecular oxygen as shown in diagram 1.

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What is Avogadro's number, and why is it useful? (3 points)
SashulF [63]

Answer:

6.022 x 10²³; it is a conversion factor between moles and number of particles

Explanation:

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole of hydrogen  = 6.022 × 10²³ atoms of hydrogen

238 g of uranium = 1 mole of uranium = 6.022 × 10²³ atoms of uranium

By taking ions:

62 g of NO⁻₃ =  1 mole of  NO⁻₃  = 6.022 × 10²³ ions of  NO⁻₃

96 g of SO₄²⁻ = 1 mole of SO₄²⁻ =  6.022 × 10²³ ions of  SO₄²⁻

4 0
4 years ago
Xavier was conducting a long term study of two populations of butterflies. Both populations had the same number of individuals.
nata0808 [166]
B. I had that same question!!
8 0
3 years ago
The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe
Irina-Kira [14]

Answer:

t=147.24\ min

Explanation:

Given that:

Half life = 30 min

t_{1/2}=\frac{\ln2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30}\ min^{-1}

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration [A_0] = 7.50 mg

Final concentration [A_t] = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

0.25=7.50e^{-0.0231\times t}

750e^{-0.0231t}=25

750e^{-0.0231t}=25

x=\frac{\ln \left(30\right)}{0.0231}

t=147.24\ min

6 0
3 years ago
Which property describes a mixture? It cannot be separated by physical methods. It has a single chemical composition. It cannot
natima [27]

Answer:

it cannot be separated by physical methods

Explanation:

7 0
3 years ago
Read 2 more answers
Complete combustion of 5.90 g of a hydrocarbon produced 18.8 g of CO2 and 6.75 g of H2O.
Inessa05 [86]
First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C 
Molar mass of H2O:
2*1+16 = 18g/mol
As there is 2 moles of H in H2O,
So,

<span>(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H </span>

<span>Divide both number of moles by the smaller number of moles: </span>
<span>As Smaaler no moles is 0.427:
So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 </span>
<span>(0.74 mol H) / 0.427 = 1.733 </span>

<span>To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula: 
C = 1 * 2 = 2
H = 1.733 * 2 =3.466
So , the empirical formula is C2H3</span>
4 0
3 years ago
Read 2 more answers
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