Answer:
∴ The absolute pressure of the air in the balloon in kPa = 102.69 kPa.
Explanation:
- We can solve this problem using the general gas law:
<em>PV = nRT</em>, where,
P is the pressure of the gas <em>(atm)</em>,
V is the volume of the gas in L <em>(V of air = 6.23 L)</em>,
n is the no. of moles of gas <em>(n of air = 0.25 mole)</em>,
R is the general gas constant <em>(R = 0.082 L.atm/mol.K)</em>,
T is the temperature of gas in K <em>(T = 35 °C + 273 = 308 K</em>).
∴ P = nRT / V = (0.25 mole)(0.082 L.atm/mol.K)(308 K) / (6.23 L) = 1.0135 atm.
- <em>Now, we should convert the pressure from (atm) to (kPa).</em>
1.0 atm → 101.325 kPa,
1.0135 atm → ??? kPa.
∴ The absolute pressure of the air in the balloon in kPa = (101.325 kPa)(1.0135 atm) / (1.0 atm) = 102.69 kPa.
How many electrons does each element have to lose again to achieve a noble gas electron configuration?
Answer:
Explanation:
Each element will gain or lose electron to attain the octet of the noble gases in their outer shell electrons.
- Most metals in group 1 and 2 will lose 1 and 2 electrons apiece to attain a noble configuration.
- Non-metals are typically electronegative and will gain considerable amount of electrons to complete their octet.
- Halogens will need one electron to complete their own out shell configuration.
Elements will gain or lose an amount of electron that will make it resemble noble gases.
Answer:
2.64575131
Explanation:
or 2.65 or 2.7
If u want it in simplest form
Hope this helped1
Have a supercalifragilisticexpialidocious day!
Answer:
B. 58°C
Explanation:
Hello,
In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:
In such a way, solving for the final temperature we obtain:
Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:
Thereby, answer is B. 58°C
.
Regards.