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3 years ago

Do molecules gain or lose energy in evaporation.

1 answer:

8 0

They gain energy. In condensation, a gas becomes a liquid when this energy is removed (the cooling process).

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Which synthetic polymer is made into fibers that do not wear out easily?

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C its Nylon that's the Synthetic polymer that doesnt wear out easily

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3 years ago

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There will be 18 oxygen Atoms

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3 years ago

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

4 years ago

Consider the following comparison.

motikmotik

Answer: B yeah I'm pretty sure

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3 years ago

To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g o

Romashka-Z-Leto [24]

Answer:

a) 1.740 g of F- must be added to a cylindrical water reservoir

b) Grams of sodium fluoride, NaF, that contain this much fluoride:

3.84 g

Explanation:

Step 1. calculate the volume of the tank:

Volume of cylinder =

$\pi r^{2}h$ ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

$r=\frac{d}{2}$

$=\frac{3.36x10^{2}}{2}$

= 168 m

Volume =

$=\frac{22\times 168^{2}\times 21.80}{7}$

$=1.93\times 10^{6} m^{3}$

2.Convert ppm to g/m3 and Solve for mass of F-

$1ppm = 1g/m^{3}$

$0.9ppm = 0.9g/m^{3}$

Because both ppm and g/m3 are same quantity .

$g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}$

$0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}$

$mass\ of\ F- =1.740g$

mass of F- required = 1.740 g

3. Apply mole concept to calculate grams of sodium fluoride produced

mass of 1 mole of F2 = 38 g

mass of 1 mole of NaF = 42 g

(from periodic table calculate molar mass)

$2Na+F_{2}\rightarrow 2NaF$

Here 1 mole of F2 produce = 2 mole of NaF

So,

38 g of F2 produce = 2 x 42 g of NaF

38 g of F2 produce = 84 g of NaF

1 g of F2 produce = 84/38 g of NaF

1.74 g F2 produce =

$\frac {84}{38}\times 1.74$

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced

3 0

3 years ago