Answer:
1 mole of platinum
Explanation:
To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.
The empirical formula for the compound can be obtained as follow:
Platinum (Pt) = 117.4 g
Carbon (C) = 28.91 g
Nitrogen (N) = 33.71 g
Divide by their molar mass
Pt = 117.4 / 195 = 0.602
C = 28.91 / 12 = 2.409
N = 33.71 / 14 = 2.408
Divide by the smallest
Pt = 0.602 / 0.602 = 1
C = 2.409 / 0.602 = 4
N = 2.408 / 0.602 = 4
The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄
From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.
13. D
14. A
15. C
16. B
17. F
18. E
Put a picture we can’t see the arrow
<span>The electron configuration that represents a violation of the pauli exclusion principle is:
</span><span>1s: ↑↓
2s: ↑↑
2p: ↑</span>
The Pauli exclusion principle refers to the quantum mechanical rule which expresses that at least two indistinguishable fermions (the particles with half-integer spin) can't involve a similar quantum state inside a quantum framework all the while.
Answer:
The temperature change from the combustion of the glucose is 6.097°C.
Explanation:
Benzoic acid;
Enthaply of combustion of benzoic acid = 3,228 kJ/mol
Mass of benzoic acid = 0.570 g
Moles of benzoic acid = 
Energy released by 0.004667 moles of benzoic acid on combustion:
Heat capacity of the calorimeter = C
Change in temperature of the calorimeter = ΔT = 2.053°C
Glucose:
Enthaply of combustion of glucose= 2,780 kJ/mol.
Mass of glucose=2.900 g
Moles of glucose = 
Energy released by the 0.016097 moles of calorimeter combustion:
Heat capacity of the calorimeter = C (calculated above)
Change in temperature of the calorimeter on combustion of glucose = ΔT'
The temperature change from the combustion of the glucose is 6.097°C.
