Answer:
the answer is mabye 30 I didnt understand
Answer:
scientist never ignore the missing -parts of body
Explanation:
they take the missing bones as an opportunity to look into them as practical for pre medical student ,they are very costly sealed
Answer:
<h2>The answer is 5000 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 1000 × 5
We have the final answer as
<h3>5000 N</h3>
Hope this helps you
Answer:
4.51 × 10³ kcal
B. less
Explanation:
Let's consider the combustion of 2-methylheptane.
C₈H₁₈ + 12.5 O₂ ⇒ 8 CO₂ + 9 H₂O
When 1 mole of C₈H₁₈ burns, 1.306 × 10³ Kcal are released (heat of combustion). The molar mass of C₈H₁₈ is 114 g/mol. Then, for 394 g of C₈H₁₈:
![394gC_{8}H_{18}.\frac{1molC_{8}H_{18}}{114gC_{8}H_{18}}.\frac{1.306 \times 10^{3} kcal }{1molC_{8}H_{18}} =4.51 \times 10^{3} kcal](https://tex.z-dn.net/?f=394gC_%7B8%7DH_%7B18%7D.%5Cfrac%7B1molC_%7B8%7DH_%7B18%7D%7D%7B114gC_%7B8%7DH_%7B18%7D%7D.%5Cfrac%7B1.306%20%5Ctimes%2010%5E%7B3%7D%20kcal%20%7D%7B1molC_%7B8%7DH_%7B18%7D%7D%20%20%3D4.51%20%5Ctimes%2010%5E%7B3%7D%20kcal)
<em>Assuming the same efficiency, would 394 grams of ethanol provide more, less, or the same amount of energy as 394 grams of 2-methylheptane?</em>
Let's consider the combustion of ethanol.
C₂H₆O + 3 O₂ ⇒ 2 CO₂ + 3 H₂O
When 1 mole of C₂H₆O burns, 326.7 Kcal are released (heat of combustion). The molar mass of C₂H₆O is 46.0 g/mol. Then, for 394 g of C₂H₆O:
![394gC_{2}H_{6}O.\frac{1molC_{2}H_{6}O}{46.0gC_{2}H_{6}O}.\frac{326.7 kcal }{1molC_{2}H_{6}O} =2.80 \times 10^{3} kcal](https://tex.z-dn.net/?f=394gC_%7B2%7DH_%7B6%7DO.%5Cfrac%7B1molC_%7B2%7DH_%7B6%7DO%7D%7B46.0gC_%7B2%7DH_%7B6%7DO%7D.%5Cfrac%7B326.7%20kcal%20%7D%7B1molC_%7B2%7DH_%7B6%7DO%7D%20%20%3D2.80%20%5Ctimes%2010%5E%7B3%7D%20kcal)
The combustion of 394 g of ethanol provides less energy than the combustion of 394 g of 2-methylheptane.
Answer:- mass of water is 0.15 g.
Solution:- This problem is based on the formula:
![q=mC_s\Delta T](https://tex.z-dn.net/?f=q%3DmC_s%5CDelta%20T)
where, q is the heat energy, m is mass,
is specific heat and
is change in temperature.
= final temperature - initial temperature
= 37.0 - 5.0 = 32.0 degree C
m = ?
q = 20.0 J
![C_s=\frac{4.184J}{g.^0C}](https://tex.z-dn.net/?f=C_s%3D%5Cfrac%7B4.184J%7D%7Bg.%5E0C%7D)
Let's plug in the values in the formula:
![20.0J=m*\frac{4.184J}{g.^0C}*32.0^0C](https://tex.z-dn.net/?f=20.0J%3Dm%2A%5Cfrac%7B4.184J%7D%7Bg.%5E0C%7D%2A32.0%5E0C)
![20.0J=m*\frac{133.888J}{g}](https://tex.z-dn.net/?f=20.0J%3Dm%2A%5Cfrac%7B133.888J%7D%7Bg%7D)
![m=20.0J(\frac{1g}{133.888J})](https://tex.z-dn.net/?f=m%3D20.0J%28%5Cfrac%7B1g%7D%7B133.888J%7D%29)
m = 0.15 g
So, the mass of water for the given problem is 0.15 g.