Answer:
<h2>━☆ﾟ.*･｡ﾟFine, but not as good</h2>
Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
Answer:
0.0611M of HNO3
Explanation:
<em>The concentration of the NaOH solution must be 0.1198M</em>
<em />
The reaction of NaOH with HNO3 is:
NaOH + HNO3 → NaNO3 + H2O
<em>1 mole of NaOH reacts per mole of HNO3.</em>
That means the moles of NaOH used in the titration are equal to moles of HNO3.
<em>Moles HNO3:</em>
12.75mL = 0.01275L * (0.1198mol / L) = 0.0015274 moles NaOH = Moles HNO3.
In 25.00mL = 0.025L -The volume of the aliquot-:
0.00153 moles HNO3 / 0.025L =
<h3> 0.0611M of HNO3</h3>
Answer : The amount of heat evolved by a reaction is, 4.81 kJ
Explanation :
Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water
![q=[q_1+q_2]](https://tex.z-dn.net/?f=q%3D%5Bq_1%2Bq_2%5D)
![q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]](https://tex.z-dn.net/?f=q%3D%5Bc_1%5Ctimes%20%5CDelta%20T%2Bm_2%5Ctimes%20c_2%5Ctimes%20%5CDelta%20T%5D)
where,
q = heat released by the reaction
= heat absorbed by the calorimeter
= heat absorbed by the water
= specific heat of calorimeter = 
= specific heat of water = 
= mass of water = 254 g
= change in temperature = 
Now put all the given values in the above formula, we get:
![q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]](https://tex.z-dn.net/?f=q%3D%5B%28783J%2F%5EoC%5Ctimes%20-2.28%5EoC%29%2B%28254g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20-2.28%5EoC%29%5D)
Therefore, the amount of heat evolved by a reaction is, 4.81 kJ
It contains nutrients and minerals that could sustain biological life
