Answer:
Explanation:
We are asked to find the specific heat capacity of a sample of lead. The formula for calculating the specific heat capacity is:
The heat absorbed (Q) is 237 Joules. The mass of the lead sample (m) is 22.7 grams. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature. The temperature increases <em>from</em> 29.8 °C <em>to </em>95.6 °C.
- ΔT = final temperature -inital temperature
- ΔT= 95.6 °C - 29.8 °C = 65.8 °C
Now we know all three variables and can substitute them into the formula.
- Q= 237 J
- m= 22.7 g
- ΔT = 65.8 °C
Solve the denominator.
- 22.7 g * 65.8 °C = 1493.66 g °C
Divide.
The original values of heat, temperature, and mass all have 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place tells us to round the 8 up to a 9.
The specific heat capacity of lead is approximately <u>0.159 Joules per gram degree Celsius.</u>
Water is the only one of these that would work by process of elimination.
When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :
when Pka = - ㏒ Ka
6.86 = -㏒ Ka
∴Ka = 1.38 x 10^-7
by using ICE table:
H2PO4- → H+ + HPO4
initial 0.4 m 0 0
change -X +X +X
Equ (0.4-X) X X
when Ka = [H+][HPO4] / [H2PO4-]
by substitution:
1.38 X 10^-7 = X^2 / (0.4-X) by solving for X
∴X = 2.3x 10^-4
∴[H+] = X = 2.3 x 10^-4
∴PH = -㏒[H+]
= -㏒ (2.3 x 10^-4)
∴PH = 3.6
Answer:
The estimated feed rate of logs is 14.3 logs/min.
Explanation:
The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.
That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.
Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.
To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.
The specific gravity of the wood chips is 0.494.
The average volume of a log is
The weight of one log is
To provide 3617 ton/day of wood chips, we need
The feed rate of logs is 14.3 logs/min.
Answer : The volume of the cube submerged in the liquid is, 29.8 mL
Explanation :
First we have to determine the mass of ice.
Formula used :
Given:
Density of ice = 
Volume of ice = 45.0 mL
The cube will float when 40.5 g of liquid is displaced.
Now we have to determine the volume of the cube is submerged in the liquid.
Thus, the volume of the cube submerged in the liquid is, 29.8 mL
