Answer:
Hi!!!
Explanation:
Hello hope you are having a great day:)
CLOUDS would be one of the interacting parts in a weather system.
If he was the primary scientist doing it as he did alot of the heavy lifting then yes its ok, but i also think how the others should also me at least mentioned. Or they could just not name the experiment by a person just so its not too biased
Answer:
1.72x10⁻⁵ g
Explanation:
To solve this problem we use the PV=nRT equation, where:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 25 °C ⇒ (25+273.16) = 298.16 K
And we <u>solve for n</u>:
- 1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
Finally we <u>convert moles of helium to grams</u>, using its <em>molar mass</em>:
- 4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g
Answer:
0.260 Celsius
Explanation:
q =c x m x (T2-T1)
c - specific heat of water 4.186 J/g.C
T2-T1 = q /(c x m) = 0.959 /(4.186 x 0.88) = 0.959/3.68 =0.260 C
