All categories

3 years ago

JOIN MY AMONG US GAME THE CODE IS QGORMF

Chemistry

2 answers:

balandron [24]3 years ago

8 0

Answer:

ill do it later

Explanation:

Andreas93 [3]3 years ago

6 0

Answer:

sure okay I'll join your game

You might be interested in

Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t

ratelena [41]

Answer:

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

$2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}$

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

$\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol$

$\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol$

$\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol$

$\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})$

$\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)$

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

The equilibrium constant determined from the partial pressure denoted as $K_p$ can be expressed as :

$K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}$

$K_p = \dfrac{1}{ (22.20)}$

$K_p$ = 0.045

$\Delta G = \Delta G^0 _{rxn} + RT \ lnK$

where;

R = gas constant = 8.314 × 10⁻³ kJ

$\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)$

$\Delta G =207.6 + 2.4788191 \times \ ln(0.045)$

$\Delta G =207.6+ (-7.687048037)$

$\Delta G =$ 199.912952 kJ

ΔG ≅ 199.91 kJ

7 0

4 years ago

The energy in eV for light with a wavelength of 6250 angstroms is _. Note - there are 1.6 x 10-12 erg in 1 eV.

vivado [14]

Answer:

2 eV

Explanation:

The energy of a photon of light is given by the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

In this problem we have:

$h=6.63\cdot 10^{-34} Js$

$c=3.0\cdot 10^8 m/s$

$\lambda=6250 A = 6250\cdot 10^{-10} m$ is the wavelength of the photon

Therefore, the energy in Joules is

$E=\frac{(6.63\cdot 10^{-34})(3.0\cdot 10^8)}{6250\cdot 10^{-10}}=3.2\cdot 10^{-19}J$

We want to convert this energy into electronvolts: we know that the conversion factor is

$1 eV = 1.6\cdot 10^{-19}J$

Therefore,

$E=\frac{3.2\cdot 10^{-19}}{1.6\cdot 10^{-19}}=2 eV$

5 0

3 years ago

Arrange the following measurements, in seconds, from the greatest bias to the least bias. 6.63 ± 0.01 s, 6.6 ± 0.1 s, 6.52 ± 0.0

lord [1]

I understand here "bias" to be the uncertainty of measurements. So the order will be the following:

6.4 ± 0.5 s
6.6 ± 0.1 s,
6.63 ± 0.01 s,
6.52 ± 0.05 s,

(notice how the second number, the one behind the symbol ± gets smaller, as the bias gets smaller).

6 0

4 years ago

How do people start alcohol consumption

Debora [2.8K]

Answer:

Explanation:Due to the mental pressure,

Due to peer pressure,

Lack of love and affection from the family members.

Influence from the T.V advertisement.

6 0

3 years ago

How many valence electrons does tungsten have

bulgar [2K]

Answer:74

Explanation:

8 0

3 years ago

Read 2 more answers