Answer:
The specific heat of zinc is 0.361 J/g°C
Explanation:
<u>Step 1:</u> Data given
44.0 J needed
Mass of solid zinc = 10.6 grams
Initial temperature = 24.9 °C
Final temperature = 36.4 °C
<u>Step 2</u>: Calculate the specific heat of zinc
Q = m*c*ΔT
⇒ with Q = heat (in Joule) = 44.0 J
⇒ with m = the mass of the solid zinc = 10.6 grams
⇒ with c = the specific heat of the zinc = TO BE DETERMINED
⇒ with ΔT = The change in temperature = T2-T1 = 36.4 °C - 24.9 °C = 11.5 °C
44.0 J = 10.6 grams * c * 11.5°C
c = 44.0 J / (10.6g * 11.5 °C)
c = 0.361 J/g°C
The specific heat of zinc is 0.361 J/g°C
Answer:
In order for the powder to dissolve, each powder molecule must separate from the other powder molecules and be surrounded by water molecules. This shift in arrangement either absorbs or releases energy depending on the situation. It is due to the exchange of energy that the temperature of the solution fluctuates.
Explanation:
:)
Explanation:
According to the Henderson-Hasselbalch equation, the relation between pH and 
is as follows.
pH = 
where, pH = 7.4 and = 7.21
As here, we can use the nearest to the desired pH.
So, 7.4 = 7.21 + 
0.19 =
= 1.55
1 mM phosphate buffer means ![[HPO_{4}]](https://tex.z-dn.net/?f=%5BHPO_%7B4%7D%5D)
+ ![[H_{2}PO_{4}]](https://tex.z-dn.net/?f=%5BH_%7B2%7DPO_%7B4%7D%5D)
= 1 mM
Therefore, the two equations will be as follows.
= 1.55 ............. (1)
+ = 1 mM ........... (2)
Now, putting the value of from equation (1) into equation (2) as follows.
1.55![[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]](https://tex.z-dn.net/?f=%5BH_%7B2%7DPO_%7B4%7D%5D%20%2B%20%5Btex%5D%5BH_%7B2%7DPO_%7B4%7D%5D)
= 1 mM
2.55 = 1 mM
= 0.392 mM
Putting the value of in equation (1) we get the following.
0.392 mM + = 1 mM
= (1 - 0.392) mM
= 0.608 mM
Thus, we can conclude that concentration of the acid must be 0.608 mM.
<u>Given:</u><u> </u>
Mass of Nitrogen (N₂) gas = 75 grams
<u>Finding the number of moles of N₂:</u><u> </u>
We know that the molar mass of N₂ is 28 grams/mole
Number of moles = Given mass / Molar mass
Number of moles = 75 / 28
Number of moles = 2.68 moles
Hence, there are 2.68 moles in 75 grams of Nitrogen Gas
