Answer:
m = 998 g
Explanation:
Hello there!
In this case, according to the definition of the molar mass as the mass of one mole of the compound, it is possible to state the 1 mole of C8H18 has a mass of 114.26 grams; therefore, the mass in 8.65 moles turn out to be:

In agreement to the notation requirement.
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Answer:
i) 0,7 molH20/s
ii)11,2 g O/s
iii)1,4 g H/s
Explanation:
i) To find the molar flow rate of water, we just convert the mass of water to moles of water using its molecular weight(g/mol) and changing to the proper units (lb to grames and hours to seconds):

ii) Now we just consider the oxygen in the water stream (for 1 mole of water there is 1 mole of oxygen):

iii)Just considering the hydrogen in the stream (for 1 mole of water there is 2 moles of hydrogen):

Answer:
Copper(II) sulphate – sodium hydroxide reaction
The reaction between copper(Il) sulphate and sodium hydroxide solutions is a good place to start. If you slowly add one to the other while stirring, you will get a precipitate of copper(II) hydroxide, Cu(OH)2.
Answer:
The answer to your question is below
Explanation:
Covalent bonds are bonds between to atoms that share a pair of electrons, there are three kinds of covalent bonds but I'll describe only two:
Covalent non polar bond: is a covalent bond between two elements of the same element. Ex two hydrogens, two chlorine, two oxygenes, etc.
Covalent polar bond: is a covalent bond between 2 elements of different elements, for example: hydrogen and chlorine or nitrogen, they are polar because on of the element that form it is smaller than the other one, then a partial positive and a partial negative charge is formed.
Answer:

Explanation:
Hello there!
Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:
![K=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.
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