Since each Chlorine molecule is -1 and wants to gain an electron, 2 Chlorine atoms like to come together to form Cl2 by sharing 2 electrons each to form a single bond between the 2 atoms. Since both Chlornine has the same electronegativity, the bond is non-polar covalent since there electrons are evenly shared.
Explanation:
Filtration is a separation technique in which solid particles suspended in liquid medium are separated by allowing the mixture through the pores of the filter paper. By this solid particles get collect on filter paper and liquid drains out from the pores of the filter paper.
The chronological order for given steps will be:
- Weigh and fold the filter paper.
- Place the filter paper in the funnel, then place the funnel in the Erlenmeyer flask.
- Allow the solid/liquid mixture to drain through the filter.
- Use water to rinse the filter paper containing the mixture.
- Weigh the dried filter paper and copper.
<span>7.379 * 10^(-4) is measured, hence prone to error, either human error or via measuring device. In this case,
100 cm = 1 m is written in stone and is unquestionable.
The density of the gold is 19.3 g/cm^3 and could be an approximation.
The approximation is good to at least one night.</span>
Answer:
Concentration of OH⁻:
1.0 × 10⁻⁹ M.
Explanation:
The following equilibrium goes on in aqueous solutions:
.
The equilibrium constant for this reaction is called the self-ionization constant of water:
![K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]](https://tex.z-dn.net/?f=K_w%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%5Ccdot%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D)
.
Note that water isn't part of this constant.
The value of 
at 25 °C is 
. How to memorize this value?
- The pH of pure water at 25 °C is 7.
![[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-%5Ctext%7BpH%7D%7D%20%3D%2010%5E%7B-7%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
- However,
![[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%3D10%5E%7B-7%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
for pure water. - As a result,
![K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}](https://tex.z-dn.net/?f=K_w%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%5Ccdot%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%2810%5E%7B-7%7D%29%5E%7B2%7D%20%3D%2010%5E%7B-14%7D)
at 25 °C.
Back to this question. ![[\text{H}^{+}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D)
is given. 25 °C implies that 
. As a result,
![\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cfrac%7BK_w%7D%7B%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%7D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B1.0%5Ctimes%2010%5E%7B-5%7D%7D%20%3D%2010%5E%7B-9%7D%20%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
Divide 1.25/3.93 and then multiply by 100 to get the percent. This equals 31.81%
