Answer:
Corrosion is the process of deterioration of materials as a result of chemical, electrochemical or other reactions. Rusting is a part of corrosion and is a chemical process which results in the formation of red or orange coating on the surface of metals. ... Rust or rusting can affect only iron and its alloys.
Explanation:
It's A. volume
Pressure =
with const depends on the chosen unit of volume
I think so...
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)
Best regards.
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
