Transferring or sharing electrons between atoms forms a covalent bond.<span> Covalent
bonding is when atoms share electrons. It is a chemical bond that involves the
sharing of electron pairs. These pairs are called bonding pairs. Examples of
compounds that has covalent bonds are CO2, organic compounds, lipids and
proteins.</span>
Answer:
By adding the solute in to solution boiling point is increased while freezing point is decreased.
Explanation:
When solute in added into the solvent the boiling point of solvent increases for example,
Water is boiled at 100 °C, when sodium chloride is added its boiling point increased. Ions of salt interact with solvent and prevent the water molecules to escape from the surface and form gas molecules. In order to make it boiled solution must be heated above 100 °C.
But there is different case with freezing point. Freezing point is the state in which substance converted into the solid. At given temperature when solute is added into the solvent it prevent the formation of solid. It required time to decrease the temperature first and as the temperature is decreases solid is formed.
Families are another names for the columns
Answer:
The new pressure is 0.5 atm
Explanation:
Step 1: Data given
Volume of oxygen = 300 mL = 0.300 L
Pressure = 1.00 atm
Temperature = 300 K
The volume increases to 1000mL = 1.00 L
The temperature increases to 500 K
Step 2: Calculate the new pressure
(P1*V1)/T1 = (P2*V2)/T2
⇒with P1 = the initial pressure = 1.00 atm
⇒with V1 = the initial volume = 0.300 L
⇒with T1 = the initial temperature = 300 K
⇒with P2 = the new pressure = TO BE DETERMINED
⇒with V2 = the increased volume = 1.00 L
⇒with T2 = the increased temperature = 500 K
(1.00 atm* 0.300 L)/300 K = (P2 * 1.00L) / 500 K
P2 = (1.00 *0.300 * 500) / (300 *1.00)
P2 = 0.5 atm
The new pressure is 0.5 atm
Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:
Now, we need to find the concentration of the OH⁻ ions.
![[OH^{-}] = \frac{n_{NaOH}}{V}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20)
Where V is the volume of the solution = 1.00 L
![[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7B3.01%20%5Ccdot%2010%5E%7B-3%7D%20moles%7D%7B1.00%20L%7D%20%3D%203.01%20%5Ccdot%2010%5E%7B-3%7D%20mol%2FL%20)
Finally, we can calculate the pH of the solution as follows:
![pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%283.01%20%5Ccdot%2010%5E%7B-3%7D%29%20%3D%202.52%20)
Therefore, the pH of the solution is 11.48.
I hope it helps you!
