Answer : The final equilibrium temperature of the water and iron is, 537.12 K
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


where,
= specific heat of iron = 560 J/(kg.K)
= specific heat of water = 4186 J/(kg.K)
= mass of iron = 825 g
= mass of water = 40 g
= final temperature of water and iron = ?
= initial temperature of iron = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:


Therefore, the final equilibrium temperature of the water and iron is, 537.12 K
Answer:
3.5 × 10⁵ g of salt
Explanation:
<em>What is the mass (grams) of salt in 10.0 m³ of ocean water?</em>
We have this data:
- 1.000 mol salt is equal to 58.44 g salt
- 1.0 L of ocean water contains 0.60 mol of salt
We will need the following relations:
We can use proportions:

Answer:
我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈
Explanation:
我實際上不知道我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈
Answer:
57.48%
Explanation:
Calculate the mass of 1 mole of malachite:
MM Cu = 63.55
MM O = 16.00
MM H = 1.01
MM C = 12.01

A mole of malachite has:
2 moles of Cu
5 moles of O
2 moles of H
1 mole of C
MW Malachite = 2*MM(CU) + 5*MM(O) + 2*MM(H) + 1 *MM(C)
MW Malachite = 2*63.55 + 5*16.00 + 2*1.01 + 1*12.01
MW Malachite = 221.13
Mass of Cu in a mole of Malachite = 2*MM(CU) = 127.1
Now divide the mass of Cu by the mass of Malachite
