Heating oxygen difluoride gas yields oxygen gas and fluorine gas? Whats the balance chemical equation for this?

Chemistry

1 answer:

N76 [4]3 years ago

5 0

2OF2(g) —> O2(g) + 2F2(g) hope that helps

You might be interested in

The half life for the decay of carbon-14 is 5.73 x 10 years. Suppose the activity due to the radioactive decay of the carbon-14

Elena-2011 [213]

Answer:

Age of the atifact is $4.2\times 10^{2}$ years

Explanation:

For first -order radioactive decay- $A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$
$A_{t}$ represents activity of radioactive nuclide after t time, $A_{0}$ represents initial activity of radioactive nuclide and $t_{\frac{1}{2}}$ represents half-life
Here, $A_{t}=19Bq$ , $A_{0}=20Bq$ and $t_{\frac{1}{2}}=5.73\times 10^{3}years$

Plug-in all the given values in the above equation-

$19=20\times (\frac{1}{2})^{\frac{t}{5.73\times 10^{3}}}$

or, $t=4.2\times 10^{2}$

So, age of the atifact is $4.2\times 10^{2}$ years

6 0

2 years ago

Why are x-rays telescopes and gamma-ray telescopes in space?

pogonyaev

Because those can leak and cause cancer

5 0

3 years ago

A sample of black coffee was measured to have [h3o+] = 7.94 × 10−6. what is true about this solution?

Sveta_85 [38]

Being given the hydronium (or simply the hydrogen) ion concentration (molarity), one of the easiest properties of the solution to determine is its pH. The pH of a solution indicates the acidity (or basicity) of a given solution. The typical pH range of solutions is from 1 to 14, with 7 being neutral, 1 to 6 being acidic, and 8 to 14 being basic.

To determine the pH of a solution, the negative log (-log) of the hydronium/hydrogen ion concentration (molarity) of the solution is calculated. Thus (assuming that the given hydrogen ion concentration is in molarity (mol/L)),

pH = -log( $H^{+}$ ) or -log( $H_{3} O^{+}$
pH = -log (<span>7.94 × 10−6 mol/L)
pH = 5.13

Because the pH is less than 7 (within the range 1 to 6), then the black coffee solution is acidic. </span>

6 0

3 years ago

Suppose protons were emitted rather than electrons. How would this affect the experiment?

Vera_Pavlovna [14]

Suppose protons were emitted rather than electrons then it affects the experiment as the mean velocity of proton will be less then mean velocity electron .

The mass of proton is greater than the mass of electrons but charge of electron is equal to the charge of proton . So , due to difference in the mass of electron and proton there will be some effects. We can conserve the electric energy which is equal to the qe .

The kinetic energy = 1/2 $mv^{2}$