M(O₂)=20g
M(O₂)=32.0 g/mol
n(O₂)=20/32.0=0.625 mol
m(C)=12 g
M(C)=12.0 g/mol
n(C)=12/12.0=1.0 mol
2C + O₂ → 2CO
1 mol 0.625 mol 1 mol
0.625-0.5=0.125 mol
2CO + O₂ → 2CO₂
0.250 mol 0.125 mol 0.250 mol
n(CO)=1 mol - 0.250 mol = 0.750 mol
M(CO)=28.0 g/mol
m(CO)=0.750*28.0=21.0 g
n(CO₂)=0.250 mol
M(CO₂)=44.0 g/mol
m(CO₂)=0.250*44.0=11.0 g
Answer:
Explanation:
Step 1. Calculate the pOH
pOH =-log[OH⁻]
pOH =-log(1.0 × 10⁻⁹)
pOH = 9.00
Step 2. Calculate the pH
pH + pOH = 14.00
pH + 9.00 = 14.00
<h3>
Answer: <em>
pH=2.25 </em></h3>
Explanation:
monochloroacetic acid also means: chloroacetic acid
pKa of monochloroacetic acid= 1.4 x 10^-3 (I believe this should have been given in the problem or perhaps in the textbook)
Formula: pH= pKa + log ( some number in M)
pH= -log (1.4 x 10^-3) + log (0.25M)= 2.85 + -0.602= 2.25
pH= 2.25
Answer: 
Explanation:
As we know that,
![-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=-%5Bm_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%5D%3D%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of metal = 100.0 g
= mass of water = 100.0 g
= final temperature = 
= temperature of metal = 
= temperature of water = 
= specific heat of metal = ?
= specific heat of water= 
Now put all the given values in equation (1), we get
![-[100.0\times c_1\times (80-600)]=[100.0\times 4.2\times (80-30)]](https://tex.z-dn.net/?f=-%5B100.0%5Ctimes%20c_1%5Ctimes%20%2880-600%29%5D%3D%5B100.0%5Ctimes%204.2%5Ctimes%20%2880-30%29%5D)
Therefore, the specific heat capacity of metal is 
Answer:
The volume of displaced fluid is equivalent to the volume of an object fully immersed in a fluid.
