Question
I need help with question 12
is How many elements exist as dialomic gases at room temperature and pressure (bl what type of tend to form cations when ionized? (c) Provide the names for the transition metals that have symbols not based on English names. the most common charge for ions of the chalcogens? True or false: Fluorine gas is the least massive How many neutrons does a 20 atom or false: A noble gas. (f atom an mass of 127.60 amu 10. Using conver discussed n class, give the name or formula as the most specific compound type (if ionic compound binary molecule appropriate for each ofthe following, and SCN or acid la (b) samous hydride, tetraphosphorus (il anic acid: cesium chromate; (k) pentane: cobaltIllI azide. l I. write out the melecular, structural and condensed structural formulas for (a) heptane, (b) 1-butanol, e)2-propanal 12. 2.50 g oro 127 mm diameter gold wire (density 19.32 gymL) cumently sells for SI284.00. using any or of the following conversions and any other metric to metric convensions, determine the cou per foot of gold in -2.54 1.057 q 1 L) Write balanced chemical equation for the complete combustion or artemisinin, cisHatos. Determine the formula weight of yttrium selenie lode comect number of significant figures letiles (el determine the sumber ef periodic table provided oxygen atoms in 600 g of yttrium selenite. 3. Determine the empirical formula of a compound that is 46.69% carbon, 53gsi hydrogen. 16.3 nia eges and 3 Lithium nitride reacts with waler to produce and lithium hydroxide (ao write balanced denial equatkon uing whole number corfficients D Determine the theoretical yield in grams of lithium can the actually isolated
Explanation / Answer
diameter d= 0.127 mm
radius r = d/2 = 0.127/2 = 0.0635 mm = 0.00635 cm ( 1mm = 0.1 cm)
1 feet of wire = 12 inches = 12 x 2.54 cm = 30.48 cm , (where 1 inch = 2.54 cm)
hence wire length l = 30.48 cm
volume of wire = (pi x r^2 x l ) ( since wire will be cylinder in shape we used this formula)
= ( 3.14 x 0.00635cm^2 x 30.48 cm )
= 0.00386 cm3
mass of wire = volume x density
= 0.00386 cm3 x 19.32 g/cm3 ( 1ml = 1cm3)
= 0.07456 g
2.5 g cost is 1284 $
0.07456 g gold cost = ( 1284/2.5) x 0.07456 = 38.29 $
