Why does pressure change in this way? Select all that apply.

If your lawn is 21.0 ft wide and 20.0 ft long, and each square foot of lawn accumulates 1350 new snow flakes every minute. How m

klemol [59]

Let us start with the total area of the lawn. Area= width x length, ie, 21 x 20 = 420 sq. ft. Snow flakes per square foot per minute = 1350 So Snow flakes for 420 sq.feet per minute = 420 x 1350 = 567000. Snow flakes for 1 hour = 567000 x 60 = 34020000 (60 minutes) Weight of 34020000 snow flakes = 34020000 x 1.60 = 54432000mg. To convert it into kilograms, divide this number by 1000000 (1 kilogram = 1000000 milligrams) Thus 54432000/1000000 = 54.432 kilograms or 54 kilograms and 432 grams.

6 0

3 years ago

The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C

Kay [80]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.775)=0.225\times 100=22.5\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

6 0

3 years ago

Which of following statements is true about scientific knowledge? Scientific ideas are not subject to change, so scientific know

Romashka [77]

Answer:

The accepted views of science knowledge can change over time. Changes can result from new science observations, but can also be affected by social, political or religious convictions. To develop a deeper understanding, students need to investigate the context of the time in which science ideas were developed.

Explanation:

7 0

3 years ago

Read 2 more answers

The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane wi

cricket20 [7]

Answer:

See explanation

Explanation:

Number of moles of each gas is

Nitrogen = 15/28 = 0.536 kmoles

Helium = 5/4 = 1.25 kmoles

Methane = 60/16 = 3.75 kmoles

Ethane = 20/30 = 0.67 kmoles

Total number of moles = 0.536 kmoles + 1.25 kmoles + 3.75 kmoles + 0.67 kmoles = 6.206 kmoles

Mole fraction of each gas;

Nitrogen = 0.536 kmoles/6.206 kmoles = 0.086

Helium = 1.25 kmoles/6.206 kmoles = 0.201

Methane = 3.75 kmoles/6.206 kmoles =0.604

Ethane = 0.67 kmoles/6.206 kmoles =0.108

Partial pressure of each gas;

Nitrogen = 0.086 * 1200 kPa = 103.2 kPa

Helium = 0.201 * 1200 kPa = 241.2 kPa

Methane = 0.604 * 1200 kPa = 724.8 kPa

Ethane = 0.108 * 1200 kPa = 129.6 kPa

Apparent specific heat at constant pressure;

Cp = (0.15 * 1.039) + (0.05 * 5.1926) + ( 0.6 * 2.2537) + (0.2 * 1.7662)

Cp = 2.12 KJ Kg-1 K-1

Cv = Cp- Ru/M

Cv= 2.12 - 8.314/16.12 = 1.604 KJ Kg-1 K-1

8 0

2 years ago

An experimenter places a piece of solid metal weighing 255g into a graduated cylinder which she then fills with mercury. After w

damaskus [11]

Answer: the density of the solid metal is 22.5176 g/cm

Explanation:

Given that;

mass of solid metal = 255 g

the cylinder and its contents weigh 101g less than before

also given that density of mercury = 13.6 g/cm³

Volume of the solid metal = volume of mercury which was absent in the first weighing

so let volume = Vcm³

Mass of mercury which was absent in the first weighing = density × volume

= 13.6 g/cm³ × Vcm³ = 13.6V g

now since the cylinder and its contents weigh 101g less than before,

it means the difference in mass between the solid and metal and the mass of mercury is equal to 101 g

so

255 - 13.6V g = 101g

13.6V g = 255 g - 101 g

13.6V = 154

V = 154 / 13.6

V = 11.3245 cm³

Now Density of solid metal will be;

⇒ mass / volume

= 255 / 11.3245

= 22.5176 g/cm

Therefore, the density of the solid metal is 22.5176 g/cm

7 0

3 years ago