Draw the Lewis structure for carbonate

Three substances A, B, and C melt at 00C,500C and -1500C

Drupady [299]

Answer:

That information is better presented and analyzed in a table.

This table shows you all the information and the answers:

Substance         melting point   boiling point    room temperature    conclusion
                                    °C                  °C                      °C                (state)

A                                  0                    100                    25                    liquid

B                                  50                  200                    25                    solid
C                               -150                   10                 25                    gas

Explanation:

1) Substance A at 25° is above the melting point and below the boiling point, then it is liquid (just like water)

2) Substance B at 25°C is below the melting point, so it is solid.

3) Substance C at 25°C is above the boiling point, so it is gas.

7 0

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

3 years ago

How many gallons of gasoline that is 5% ethanol must be added to 2,000 gallons of gasoline with no ethanol to get a mixture that

Amanda [17]

Answer:

$V_1= 3000 gal$

Explanation:

We have 3 solutions:

Solution 1 (with ethanol)
Solution 2 (no ethanol)
Final solution

$V_f=V_1 + V_2$

and for the ethanol:

$V_f*0.03=V_1*0.05 + V_2*0$

$V_f=V_1 \frac{5}{3}$

Combining:

$V_1 \frac{5}{3}=V_1 + V_2$

$V_1 \frac{2}{3}= V_2$

$V_1= \frac{3}{2} V_2$

If V2=2000 gal:

$V_1= \frac{3}{2} 2000 gal$

$V_1= 3000 gal$

8 0

3 years ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is: