Is drawing on paper a chemical or physical change

Balance CH4+O2->CO2+H2O

malfutka [58]

Explanation:

I hope this helps Chemistry is so hard and I hate it

5 0

3 years ago

Given the following UNBALANCED reaction: NH3 (g) <--> N2 (g) + H2 (g) If 1

Yakvenalex [24]

Answer:

C. 1.35

Explanation:

2NH3 (g) <--> N2 (g) + 3H2 (g)

Initial concentration 2.2 mol/0.95L 1.1 mol/0.95L 0

change in concentration 2x x 3x

-0.84 M +0.42M +1.26M

Equilibrium 1.4 mol/0.95L=1.47M 1.58 M 1.26 M

concentration

Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M

Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M

Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M

K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M

8 0

2 years ago

The mass spectra of alcohols often fail to exhibit detectable M peaks but instead show relatively large ________ peaks.

nirvana33 [79]

Answer: M-18

Explanation:

The mass spectra of alcohols often fail to exhibit detectable M peaks but instead show relatively large __M-18___ peaks.

Mass spectroscopy is used to determine the molecular mass and molecular formula of a sample.

When the mass spectra of alcohols do not show detectable M peaks, they show relatively large M-18 peaks.

6 0

3 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

3 years ago

If the density of water is 1.00 g/mL and the density of mercury is 13.6 g/mL, how high a column of water in meters can be suppor

dexar [7]

Answer:

The answer to this is

The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters

Explanation:

To solve this we first list out the variables thus

Density of the water = 1.00 g/mL =1000 kg/m³

density of mercury = 13.6 g/mL = 13600 kg/m³

Standard atmospheric pressure = 760 mmHg or 101.325 kilopascals

Therefore from the equation for denstity we have

Density = mass/volume

Pressure = Force/Area and for a column of water, pressure = Density × gravity×height

Therefore where standard atmospheric pressure = 760 mmHg we have for Standard tmospheric pressure= 13600 kg/m³ × 9.81 m/s² × 0.76 m = 101396.16 Pa

This value of pressure should be supported by the column of water as follows

Pressure = 101396.16 Pa = kg/m³×9.81 m/s² ×h

∴ $h = \frac{101396.16}{(1000)(9.81)}$ = 10.336 meters

The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters

4 0

3 years ago