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Answer:

Explanation:
Hello,
In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

We can represent the heats in terms of mass, heat capacities and temperatures:

Thus, we solve for the mass of platinum:

Next, by using the density of platinum we compute the volume:

Which computed in terms of the edge length is:

Therefore, the edge length turns out:
![a=\sqrt[3]{180cm^3}\\ \\a=5.65cm](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B180cm%5E3%7D%5C%5C%20%5C%5Ca%3D5.65cm)
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Explanation:
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Answer: -
Mass of Hydrated KAl(SO₄)₂ = 2.0 g
Molar mass of anhydrous KAl(SO₄)₂ = 258.20 g/ mol
Mass of of anhydrous KAl(SO₄)₂ = mass of the 2nd heating = A
The mass of water released = mass of the Aluminum Cup + 2.0 grams of KAl(SO₄)₂ - mass of the 2nd heating
= H g
Moles of water released = 
Moles of anhydrous KAl(SO₄)₂ = 
Required ratio = 
Answer:
There are 198 grams of iron in a 450-gram sample of the ore.
Explanation:
From the given information:
Given that:
100-gram sample of iron ore is found in 44g of iron.
Thus. the grams of iron that can be found in a 450-gram sample of the iron ore can be computed as follows:
100g of iron ore = 44g of iron
450 gram of iron ore = x g of iron
x(g) of iron = ( 44g of iron × 450 gram of iron ore) / 100g of iron ore
x(g) of iron = 198 grams of iron
Thus, there are 198 grams of iron in a 450-gram sample of the ore.