When 1.375 g of copper (II) oxide is reduced on heating in a current of hydrogen, the weight of copper remaining after the react
1 answer:
Answer:
20.11% of oxygen
Explanation:
The reaction is:
2CuO ⇒ 2Cu + O2
If you have an initial amoung of a sample, thi may be have impurities, so you have to start from products to know what you had at the begining.
to make a relation between reagents and products you need to change the mass to mol, then you can relate them with the molar ratio.
Now you have the initial amount of oxygen, to convert to percent you must divide the mass of oxygen between the mass of the sample and multiply by 100
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Answer:
Explanation:
Hello.
In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:
Thus, in terms of masses, specific heats and temperatures we can write:
Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:
Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:
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