I think it might be lead? I'm not quite too sure though. You might have to research more into it.
Answer: The mass of aluminium sample is 55.4 gram
Explanation:
As we know that,
![m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
where,
= mass of aluminium = ?
= mass of water = 300.0 g
= final temperature = 
= temperature of aluminium = 
= temperature of water = 
= specific heat of aluminium = 
= specific heat of water = 
Now put all the given values in equation (1), we get
![-[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)
![-[m_1\times 0.897\times (23.8-94.5)^0C]=[300.0g\times 4.184\times (23.8-21.0)]](https://tex.z-dn.net/?f=-%5Bm_1%5Ctimes%200.897%5Ctimes%20%2823.8-94.5%29%5E0C%5D%3D%5B300.0g%5Ctimes%204.184%5Ctimes%20%2823.8-21.0%29%5D)
Therefore, the mass of aluminium sample is 55.4 gram
C hydroyen
its aht plants breath
<span>If a lever has a constant force applied at 90∘ to the lever arm at a fixed distance from the pivot point (point of rotation), the torque on the lever is given by Torque = force (at 90∘ to lever arm) × distance to pivot point
Force should be perpendicular to the lever arm
Therefore; you would do 15N X 2.0 m = 30 Nm or 30Joules</span>
Answer:
Explanation:
First we need to find how many moles of ammonium weigh 62 grams.
Molar mass of NH4 = (14.0)+(4*1.0) grams
or 18.0 grams/mole
62 (g)/18(g/mole) = 3.444... moles of NH4
If it is dissolved in 5 litres of water, the concentration will be 3.444moles/5L
or 0.6888 M.
