All categories

2 years ago

Which of the following is not a step in the conversion of 25 L to hm3?

1 answer:

3 0

To convert a value from one unit to another, we need a conversion factor to multiply with the value. For this case, we know that for 1hm it is equal to 100 m and that 1 m^3 is equal to 1000 L. We use these:

25 L (1 m^3 / 1000 L ) ( 1 hm / 100 m )^3 = 2.5x10^-8 hm^3

Therefore, the correct answer should be the first option. The statement "fraction cubed: one m over ten to the negative second power m" is not a correct factor.

You might be interested in

Roundup, an herbicide manufactured by Monsanto, has the formula C₃H₈NO₅P. How many moles of molecules are there in a 304.3-g sam

jekas [21]

The molar mass is 169.09

304.3g/169.09g = 1.799mol which rounds to 1.800 mol

7 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

On a summer day, you take a road trip through Death Valley, California, in an antique car. You start out at a temperature of 21°

MrRissso [65]

There is only one formula to use and we should assume ideal gas. This equation is: PV=nRT. For the following questions manipulate this equation to get the answer.
1. n = PV/RT = (249*1000 Pa)(15.6 L)(1 m^3/1000 L)/(8.314 Pa-m^3/mol-K))(21+273) = 1.59 mol
2. P = nRT/V = (1.59)(8.314)(51+273)/(15.6/1000)(1000) = 274.55 kPa
3. Since the answer in #2 is more than 269 kPa, then the tires will likely burst. 4. Reduce pressure way below the limit 269 kPa.

4 0

3 years ago

SCIENCE HELP! 10 POINTS

Zepler [3.9K]

We are not 8n your. lass we need mor detail??

6 0

3 years ago

204 g of Thallium & 35 g of chlorine?

r-ruslan [8.4K]

204Tl, syn, 3.78 y, β · Pb · ε · Hg. 205Tl, 70.5%, stable. Category: Thallium · view · talk · edit. | references. Thallium is a chemical element with the symbol Tl and atomic number 81. It is a gray ... The radioisotope thallium ...

6 0

2 years ago