Convergent boundaries are zones on what plates

A sample of CH4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is 98.70

kakasveta [241]

Explanation:

The given data is as follows.

$P_{atm}$ = 98.70 kPa = 98700 Pa,

T = $30^{o}C$ = (30 + 273) K = 303 K

height (h) = 30 mm = 0.03 m (as 1 m = 100 mm)

Density = 13.534 g/mL = $13.534 g/mL \times \frac{10^{6}cm^{3}}{1 m^{3}} \times \frac{1 kg}{1000 g}$

= 13534 $kg/m^{3}$

The relation between pressure and atmospheric pressure is as follows.

P = $P_{atm} + \rho gh$

Putting the given values into the above formula as follows.

P = $P_{atm} + \rho gh$

= $98700 Pa + 13534 \times 9.81 \times 0.03 m$

= 102683.05 Pa

= 102.68 kPa

thus, we can conclude that the pressure of the given methane gas is 102.68 kPa.

8 0

4 years ago

Chlorine has 7 valence electrons. With what group of elements will chlorine most readily bond?

scZoUnD [109]

Answer:

group 1

Explanation:

chlorine donate electron

8 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

2. Name three factors that influence the rate at which a solute dissolves in a solvent?

agasfer [191]

Answer:

1. Temperature

2. Surface area

3. Catalyst

Explanation:

3 0

3 years ago

Read 2 more answers

A nonzero net force is called a(n) _____ force.

MrRa [10]

I believe it is called a non-zero net force

7 0

3 years ago

Read 2 more answers