1 mL = L
so $155$ mL =$155\times10^{-3}$ L
Now, $1$ gallon = $3.785$ L , or $1$ L= $\frac1{3.785}$ gallons
so $155\times10^{-3}$ L= $155\times10^{-3}\times \frac{1}{3.785}=40.95\times 10^{-3}$ gallons
and this is for each day, so just divide by 1 (day) to get the ratw gallons/day
i.e. $40.95\times 10^{-3}$ gal/day
What mass of water is needed to dissolve 34.8 g of copper(II) sulfate in order to prepare a 0.521 m solution? Calculate the freezing point of a solution made from 32.7 g of propane, C3H8, dissolved in 137.0 g of benzene, C6H6. The freezing point of benzene is 5.50° C and its Kf is 5.12° C/m.\ Calculate the concentration of nitrogen gas in a 1.00 L container exerting a partial pressure of 572 mm Hg at room temperature. Henry’s law constant for nitrogen at 25° C is 6.8 x 10-4 mol/L·atm.
Answer:
The number 10,847,100 in Scientific Notation is
Explanation:
Scientific notation is an easy form to write long numbers and it is commonly used in the scientific field. To write a long number in a shorter way it is necessary to 'move' the decimal point to the left the number of positions that are necessary until you get a unit. Then you write the number and multiplied it by 10 raised to the number of positions you moved the decimal point. In this case, it is necessary to move the decimal point 7 positions so, we multiply the number by 10 raised to 7.
Answer:
T₂ = 687.6 K
Explanation:
Given data:
Initial pressure = 108 Kpa
Initial temperature = 20°C
Final temperature = ?
Final pressure = 2.50 atm
Solution:
Initial pressure = 108 Kpa = 108/101 = 1.06588 atm
Initial temperature = 20°C = 20+273.15 = 293.15 k
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
1.06588 atm / 293.15 K = 2.5 atm/T₂
T₂ = 2.5 atm ×293.15 K / 1.06588 atm
T₂ = 732.875 atm. K /1.06588 atm
T₂ = 687.6 K