Answer:
221 °C
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 4.1 L
Initial temperature (T₁) = 25 °C
= 25 °C + 273
= 298 K
Final volume (V₂) = 6.8 L
Final temperature (T₂) =?
The final temperature of the gas can be obtained as follow:
V₁ / T₁ = V₂ / T₂
4.1 / 298 = 6.8 / T₂
Cross multiply
4.1 × T₂ = 298 × 6.8
4.1 × T₂ = 2026.4
Divide both side by 4.1
T₂ = 2026.4 / 4.1
T₂ ≈ 494 K
Finally, we shall convert 494 K to celcius temperature. This can be obtained as follow:
°C = K – 273
K = 494
°C = 494 – 273
°C = 221 °C
Thus the final temperature of the gas is 221 °C
Answer:
0.209M
Explanation:
M1V1=M2V2
(28.5 mL)(0.183M)=(25.0mL)(M)
M2= 0.209M
*Text me at 561-400-5105 for private tutoring if interested: I can do homework, labs, and other assignments :)
Answer:
<em>Zinc nitrate is an inorganic chemical compound with the formula Zn(NO3)2. This white, crystalline salt is highly deliquescent and is typically encountered as a hexahydrate Zn(NO3)2•6H2O. It is soluble in both water and alcohol.</em>
Explanation:
correct me if im wrong please
Sorry If This Is Late***
<u>Elements can't be broken down</u>, into a simplre set of properties. They are one strict unit & cannot be broken down, however they can be added together to make a compound.
HOPE THIS HELPS & good luck <3 !!!!!
