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3 years ago

A possible answer to a scientific question is known as

Chemistry

1 answer:

posledela3 years ago

7 0

A hypothesis, or educated guess, is a possible answer to a scientific question.

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A gas under a pressure of 74 mmHg and at a temperature of 75°C occupies a 500.0-L container. How many moles of gas are in the co

Leona [35]

Answer : The number of moles of gas present in container are 1.697 mole.

Explanation :

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of gas = 74 mmHg = 0.097 atm

conversion used : (1 atm = 760 mmHg)

V = volume of gas = 500.0 L

T = temperature of gas = $75^oC=75+273=348K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get the number of moles of gas in the container.

$(0.097atm)\times (500.0L)=n\times (0.0821L.atm/mole.K)\times (348K)$

$n=1.697mole$

Therefore, the number of moles of gas present in container are 1.697 mole.

4 0

3 years ago

Plates that are floating on molten<br> 5. The Earth's surface is made of<br> rock.

irinina [24]

What's the question?

8 0

4 years ago

Assuming a car (with a 70-L) gas tank can hold approximately 50,000 (5.00 * 10^4) g of octane(C8H18) or 50,000 (5.00 * 10^4) g o

konstantin123 [22]

Answer:

- From octane: $m_{CO_2}=1.54x10^5gCO_2$

- From ethanol: $m_{CO_2}=9.57x10^4gCO_2$

Explanation:

Hello,

At first, for the combustion of octane, the following chemical reaction is carried out:

$C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O$

Thus, the produced mass of carbon dioxide is:

$m_{CO_2}=5.00x10^4gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{8molCO_2}{1molC_8H_{18}}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=1.54x10^5gCO_2$

Now, for ethanol:

$C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O$

$m_{CO_2}=5.00x10^4gC_2H_6O*\frac{1molC_2H_6O}{46gC_2H_6O}*\frac{2molCO_2}{1molC_2H_6O}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=9.57x10^4gCO_2$

Best regards.

3 0

3 years ago

Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0

3 years ago

Match the part of the wave to the diagram

erik [133]

Answer:

The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5.

Explanation:

The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5 because of the diagram.

5 0

2 years ago