Question

CHEM QUESTIONS PLEASE ANSWER CORRECTLY !! AND EXAMPLE HOW TO DO IT

1.Classify each of the following as a chemical change or physical change: burning gasoline, making ice cubes, boiling oil, melting lead, rusting iron, making ammonia from nitrogen and hydrogen, digesting food

2. how many grams are in the following? a. 1kg b.1mg

3. make the following conversions

a. 42.6 kg to L

B . 1.62 lb to g

c.34 in to cm

d.37.2 km to mi

e.2.73 gal to L

f. 62 g to oz

g. 33.61 qt to L

h. 42.7 L to gal

i. 1.1 mi to km

j. 34.9 mL to fl oz

4. make the following metric conversions (plz explain how)

a. 96.4 mL to L

b. 275mm to cm

c. 457. kg to

G d.475 cm to m

e. 21.64 cc to mL

f. 3.29 L to CC

g.0.044 L to mL

h. 711 g to kg

i. 63.7 mL to cc

j. 0.073 kg to mg

k.83.4 m to MM

i. 361 mg to G

Answer 1

Water dioxide Solution (1) Burning of gasoline is a chama chemical change, because of gasoline is bumed in the presence of oxygen carbon dioxide and water is formed'. Ce Hie + 12.50, —> 8002 + 9H20 gasoline Oxygen Carbon Burning of Casoline — Chemical change. Makine ice cubes is a physical change, because there is only change in the physical state not its chemical composition H2O lice) - Hy water) Therefore, making of ice cubes physical change. Boiling of oil is also a physical, since ment there is no change in its chemical composition. Therefore, Boiling of oil - physical change Melting of lead physical change. Rusting of iron is a chemical change, since iron (Fe) reacts with moisture to give yust aFe 203). . Therefore, Rusting of iron chemical change Making of ammonia from nitrogen and hydrogen is a chemical ellenge ens is described by the following chemical N2 + 3H2 - -> 2NH2 Nitrogen Hydrogen .. Ammonia "Therefore, Marting of ammonia – chemical change. equation

Digesting food -> chemical change Conclusion - Chemical Change Physical Change. (0) Buoning of gasoline (2) Making of ice cubes (21) Rusting of aron. [(at) Boiling of oil. Giul Making of ammonia from c) Melting of leads nitrogen and hydrogen. (av) Digesting food. (2) How many grams are in the following: aj 1 kg = 1000 grams or 1000g : (b) img = 0.001 grams or 0.001g. (3)→ Make the following conwersions : (a) 42.6 kg to L To convert kg , le, mais to Live, volume , we need - And Density = man And for water density = 1 kg/l os i = mars = massa volume. - Volume This means 42.6 lg = 42.6L (6) 1.62 éb to go Note: Ilb = 453.59 a. So, we can write. 1.62x1 lb = lo QX 453.590. 1.62 lb = 7341829. density.

(c) 34 in to emang Note: In = 2.54cm , So, we can write guy in 34%2:54 cm 34 in 26-36 cm. (d) 37.2 km lo mi Note: 1 km = 0.62 mi So, we can write . 21.2 x 1 kom - 21.270.62 mil 31.2 mm = 22.11 mi (e) 2:13 gal to L --- Note: I gal = 3.182 So, we com write 1341 gol = 9.7243:18L gal = 10:22 1- (F) 629 to 02 --> Note: 19 = 0.035 02 So, we can write 62x19 = 62 %0.0359 G2g = 218 o2 (9) 33.61 qt to 2 - Note: lat = 100 kg and 1 kg = 1L, - so that lat = 100L we can write 33•61x1qt - 3361 %100 L 38.61 98 - 3961 L (h) 42.7L togal Note ; IL '= 0.26 gal So, we can write 0 4271L = 42.7%0.26 aal. 429L - 11. 28 al 0 (i) 1.1 mi to km Note: I m = 1.609 km So, we can write 1x1 mi - 1.1x1.609 Icm hol mi - 177 km

1.;) 34.9 ml to floz -- Note: 1 ml = 0.0338 f1 02 So, we can write. 34.9x1mk - 34.90.0338 floz 34.9mL = 1.18 floz 14) Make the following metric conversions (a) 96.44 ml to L Note : Iml = 0.001L So, we can write 96.4x1m2 = 96.4 %0.001L 96.comL = 0.0964 L (b) 275 mm to cm → Note : Lom to Imm = 0*1 cm So, we can write 75 x 1mm = 215 xolcm 275 mm = 27.5 cm (C) 457 kg to 9 → Note : Ileg 1000g So, we can write 0 457x 1 kg = 457x10009 4571g (= 4570004 (d) 475 cm to m -> Note: Tem - 0.0lm So, we can write 475x1cm = 475*0.0lm 475cm = 4.75m

(e) 21.64 cc bomla Note: Icc = Iml So, we can write 21.64x1cc = 21.64% 1ml 21.64 cc = 21.64 mL (f) 3.29 L to ce → Note: IL - 1000 cc So, we can write 3.29x1L - 3.24 x 1000 cc 3.29 L: 3290 cc (g) 0.044 L to ml → Note : Il = 1000 mL So, we can write 0.044x1L - 0.044 x1000 m2 0.044 L = 44 m2 (n) Til to kg Note: 19 = 09001 kg So, we lan write o 711x19 = t1lX0.001 kg. lg & 0.711 kg → (i) 63.7 mL to co Note: Im Lalec So, we can write 63.7 x 1 ml = 63.7xloc 63.7mL = 63.7cc (j) 0.073 kg to me -> Note: 1 kg = 1,000,000 mg So, we esin write 0.083x1 kg = 0.073X1000000 me - 0.073 10g = 13000 mg (k) 83.4 m to mm Note : Im = 1000 mm So, we can write 83.471m = 8314*1000 mm 83-4m 83400 mm

(i) 861 mg to g Note: I'mg = 0.001g So, we can write " 361x1mg = 361 *0.001 g 361 mg' = 0.3619