Answer:
New volume of gas = 95.93 ml (Approx)
Explanation:
Given:
Old volume of gas = 86 ml
Old temperature = 30°C = 30 + 273 = 303 K
New temperature = 65°C = 65 + 273 = 338 K
Find:
New volume of gas
Computation:
V1T2 = V2T1
(86)(338) = (V2)(303)
New volume of gas = 95.93 ml (Approx)
The formula of the ppt. formed is PbSo4 , which is inslouble.
The Oxyfuel gas or flame refers to a group of welding processes that use the flame produced by the combination of a fuel gas and oxygen as the source of heat.
<u>Explanation:</u>
- Oxy-fuel welding is a process that utilizes fuel gases and oxygen to weld metals. Oxyfuel gas or flame refers to a group of welding processes that utilize the flame delivered by the blending of fuel gas and oxygen as the source of heat.
- This flame is utilized for cutting and welding of two metallic pieces. This is done due to the heat produced by cutting and welding of two metallic pieces together by heating to the melting point.
- An oxyhydrogen flame is utilized for cutting and welding of two metallic pieces due to the heat produced by the flame, i.e, 2800 ° C. At this temperature, the metal gets softened effectively and thus it can easily separate or welded together.
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.
