1) number of moles of N2 = n/2
2) Number of moles of CH4 = n/2
3) Total number of moles of the mixture = n/2 + n/2 = n
4) Kg of N2
mass in grams = number of moles * molar mass
molar mass of N2 = 2 * 14.0 g/mol = 28 g/mol
=> mass of N2 in grams = (n/2) * 28 = 14n
mass of N2 in Kg = mass of N2 in grams * [1 kg / 1000g] = 14n/1000 kg = 0.014n kg
Answer: mass of N2 in kg = 0.014n kg
Answer: If you smell a dusty or burning smell the first few times you turn on your heat, it’s most likely dust and dirt that’s settled on components inside your heating system throughout the summer. As you fire up the heat, those dust particles burn away, producing a weird burnt/dusty smell.
Explanation: please mark brainlyest
Here we have to get the spin of the other electron present in a orbital which already have an electron which has clockwise spin.
The electron will have anti-clockwise notation.
We know from the Pauli exclusion principle, no two electrons in an atom can have all the four quantum numbers i.e. principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s) same. The importance of the principle also restrict the possible number of electrons may be present in a particular orbital.
Let assume for an 1s orbital the possible values of four quantum numbers are n = 1, l = 0, m = 0 and s = 
.
The exclusion principle at once tells us that there may be only two unique sets of these quantum numbers:
1, 0, 0, + and 1, 0, 0, -.
Thus if one electron in an orbital has clockwise spin the other electron will must be have anti-clockwise spin.
Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
