`0. 04xx0. 0162` is equal to `6. 48xx10^(-3)` (b) `6. 48xx10^(-4)` (c) `6. 48xx10^(-5)` (d) `6. 48xx10^(-6)`

0.1xx0.01xx0.001xx10^6 is equal to:

10xx10^(3)*2.2xx10^(-6)*(10)/(0.4)

Calculate the following a. (6.7 xx 10^(5)) xx (4.6 xx 10^(4)) b. (7.6 xx 10^(7)) xx (3.8 xx 10^(-4)) c. (6.8 xx 10^(-3)) xx (5.2 xx 10^(-4)) d. (6.7 xx 10^(5)) / (4.6 xx 10^(4)) e. (7.6 xx 10^(7)) / (3.8 xx 10^(-4)) f. (6.8 xx 10^(-3)) / (3.8 xx 10^(-4)) g. 7.65 xx 10^(2) + 2.72 xx 10^(3) h. 7.87 xx 10^(-4) - 2.61 xx 10^(-5)

If sqrt(.04xx.4xx a)=.004xx.4xx sqrt(b), then (a)/(b) is 16xx10^(-3)(b)16xx10^(-4)(c)16xx10^(-5) (d) None of these

Henry 's law constant for the solubility of N_(2) gas in water at 298 K is 1.0 xx 10^(5) atm . The mole fraction of N_(2) in air is 0.6 . The number of moles of N_(2) from air dissolved in 10 moles of water at 298 K and 5 atm pressure is a. 3.0 xx 10^(-4) b. 4.0 xx 10^(-5) c. 5.0 xx 10^(-4) d. 6.0 xx 10^(-6)

A negatively charged particle called Negatron was discovered. In the Millikan's oil-drop experiment, the charges of the oil-drops in five experimetns are reported as 3.2xx10^(-19) coulomb, 4.8xx10^(-19) coulomb, 6.4xx10^(-19) coulomb 8xx10^(-19) coulomb and 9.6xx10^(-19) coulomb. Calculate the charge on the negatron.

0.000006723 when expressed in scientific notation,is 6723xx10^(-5) (b) 67.23xx10^(-7) (c) 6.723xx10^(-6)(d) None of these