Answer:
The final temperature of the water is 28.98 degree Celsius.
Explanation:
It is given that,
Mass of sample of water, m = 52 grams
Initial temperature, ![T_i=10^{\circ}C](https://tex.z-dn.net/?f=T_i%3D10%5E%7B%5Ccirc%7DC)
Heat absorbed, ![Q=4,130\ J](https://tex.z-dn.net/?f=Q%3D4%2C130%5C%20J)
The specific heat of water is ![4.184\ J/(g^{\circ} C)](https://tex.z-dn.net/?f=4.184%5C%20J%2F%28g%5E%7B%5Ccirc%7D%20C%29)
We need to find the final temperature of the water. The heat absorbed is given by the formula as follows :
![Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\T_f=\dfrac{Q}{mc}+T_i\\\\T_f=\dfrac{4130}{52\times 4.184 }+10\\\\T_f=28.98^{\circ} C](https://tex.z-dn.net/?f=Q%3Dmc%5CDelta%20T%5C%5C%5C%5CQ%3Dmc%28T_f-T_i%29%5C%5C%5C%5CT_f%3D%5Cdfrac%7BQ%7D%7Bmc%7D%2BT_i%5C%5C%5C%5CT_f%3D%5Cdfrac%7B4130%7D%7B52%5Ctimes%204.184%20%7D%2B10%5C%5C%5C%5CT_f%3D28.98%5E%7B%5Ccirc%7D%20C)
So, the final temperature of the water is 28.98 degree Celsius.
Answer:
2 to 1
Explanation:
The reaction equation is given as:
2KHCO₃ → K₂CO₃ + CO₂ + H₂O
The conversion factor between the moles of KHCO₃ and K₂CO₃ is 2 to 1.
From the balanced chemical equation, we find out that;
2 moles of 2KHCO₃ will produce 1 mole of K₂CO₃.
So given any mole of either of the reactant or product, we can find the unknown.
Answer:
The value of
for this reaction at 1200 K is 4.066.
Explanation:
Partial pressure of water vapor at equilibrium = ![p^o_{H_2O}=15.0 Torr](https://tex.z-dn.net/?f=p%5Eo_%7BH_2O%7D%3D15.0%20Torr)
Partial pressure of hydrogen gas at equilibrium = ![p^o_{H_2}=?](https://tex.z-dn.net/?f=p%5Eo_%7BH_2%7D%3D%3F)
Total pressure of the system at equilibrium P = 36.3 Torr
Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:
![P=p^o_{H_2O}+p^o_{H_2}](https://tex.z-dn.net/?f=P%3Dp%5Eo_%7BH_2O%7D%2Bp%5Eo_%7BH_2%7D)
![p^o_{H_2}=P-p^o_{H_2O}=36.3 Torr- 15.0 Torr = 21.3 Torr](https://tex.z-dn.net/?f=p%5Eo_%7BH_2%7D%3DP-p%5Eo_%7BH_2O%7D%3D36.3%20Torr-%2015.0%20Torr%20%3D%2021.3%20Torr)
![3 Fe(s) 4 H_2O(g)\rightleftharpoons Fe_3O_4(s) 4 H_2(g)](https://tex.z-dn.net/?f=3%20Fe%28s%29%204%20H_2O%28g%29%5Crightleftharpoons%20Fe_3O_4%28s%29%204%20H_2%28g%29)
The expression of
is given by:
![K_p=\frac{(p^o_{H_2})^4}{(p^o_{H_2O})^4}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%28p%5Eo_%7BH_2%7D%29%5E4%7D%7B%28p%5Eo_%7BH_2O%7D%29%5E4%7D)
![K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}=4.066](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%2821.3%20Torr%29%5E4%7D%7B%2815.0%20Torr%29%5E4%7D%3D4.066)
The value of
for this reaction at 1200 K is 4.066.
Answer:
a) 300 L
b) 100 L
c) 12000 L
Explanation:
a) Amount of water required for 1 kg of Paper = 300 - 400 Kgs or liters
Amount of water water required for 1 Kg of steel = 704.9 L
It takes 300 L more water
b) Amount of water required for 1 kg Potato = 300 liters
It takes 100 L more water
c) Amount of water required for 1 kg Potato = 300 liters
Amount of water required for 1 kg beef = 15,415 liters
It takes 12000 L more water
1L = 33.814 oz
xL = 2.75 oz
so it's a proportion
1L / 33.814 oz = xL / 2.75
solve for x
(1/33.814) * 2.75 = 0.0813272609 on your calculator, but it's not the answer.
the number in your problem, 2.75 oz, has 3 significant figures. so you can only round this number to 3 significant figures too.
your equipment isn't accurate enough to give a reading to 10 significant figures if that makes sense. you have to give the answer in terms of the term you use with the lowest significant figures.
so with 3 significant figures,
0.0813272609 rounds to
0.0813 L