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2 years ago

A student collected the following data for a fixed volume of gas:

1 answer:

3 0

First, you should convert the temperature unit to absolute temperature.

Second, you shoul graph the points. Then you will find a pretty linear correlations among the points.

You can pick between using the best fit line or you could observe that as you get to higher temperatures the linear behavior is "more perfect".

I found this best fit line:

P = 2.608T + 14
Then, for T = 423K

P = 2.608(423) + 14 = 1117 mmHg

If you prefer to use the last two points, this is the calculus:

[P - P1] / [T - T1] = [P2 - P1] / [T2 - T1]

[P - 960]/[423 -373] = [960 - 880] / [373- 343]

=> P = 1093.3 mmHg.

You can pick any of the results 1177 mmHg or 1093 mmHg, You need more insight to choose one of them: conditions and error of the experiment for example.

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ASAP, NO TROLLS 20 points!!!!

MA_775_DIABLO [31]

Answer:

1.- Chemical change

2.- Because the atoms in the image are tranformed into a new molecule

3.- Yes, it does

4.- Due to the amount of atoms are the same both in products and reagents

Explanation:

6 0

2 years ago

Read 2 more answers

How many neutron and protons are in lithium-7?

ikadub [295]

So it contains three protons and four neutrons

6 0

3 years ago

How do scientists determine locations for drilling for natural gas? Where are many natural gas drill sites located? (Site 2)

Alchen [17]

Explanation:

By drilling several cores miles apart , geologist can correlate the rock units and create an image what exists and where below the surface.

Combining this information with rocks exposed to the surface , which can gives clues about rock orientation below, can be a powerful tool to locating oil and gas .

5 0

2 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

2 years ago

Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2

Flauer [41]

Answer:

$\boxed{28.81}$

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 58.12 44.01

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g: 9.511

1. Moles of C₄H₁₀

$\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}$

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

$\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}$

3. Mass of CO₂

$\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}$

8 0

2 years ago