That statement is true because to make a diluted solution, you have to add more solvent than the solute in a concentrated solution. This is extra information, but to make a diluted solution more concentrated, you have to add more solute since a concentrated solution has an excess amount of solute as stated before.
Answer: A barrier should be created to overcome the atmosphere of the Venus, while launching spacecraft to Venus.
Explanation:
The atmosphere of Venus consists of 96.5% carbon dioxide, other composition includes nitrogen and other gases in trace amounts. The large amount of carbon dioxide in the atmosphere can extinguish the missile of the launcher of spacecraft thus it will become difficult in launch of spacecraft to the Venus.
Answer: hydrogen
Explanation: hydrogen gas is a major component of water which occupies a large portion of the Earth's atmosphere
Answer:
![[H^{+}] = 0.761 \frac{mol}{L}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%200.761%20%5Cfrac%7Bmol%7D%7BL%7D)
![[OH^{-}]=1.33X10^{-14}\frac{mol}{L}](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D1.33X10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7BL%7D)

Explanation:
HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.
. V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)](https://tex.z-dn.net/?f=n_%7BH%5E%7B%2B%7D%20%7D%20from%20HCl%20%3D%20%5BHCl%5D%28%5Cfrac%7Bmol%7D%7BL%7D%29.%20V_%7BHCl%7D%28L%29%20%20%5C%5C%20n_%7BH%5E%7B%2B%7D%20%7D%20from%20HNO_%7B3%7D%20%20%3D%20%5BHNO_%7B3%7D%5D%28%5Cfrac%7Bmol%7D%7BL%7D%29.%20V_%7BHNO_%7B3%7D%7D%28L%29)
Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows






For molar concentration of hydrogen ions:
![[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%20%3D%20%5Cfrac%7Bn_%7BH%5E%7B%2B%7D%7D%28mol%29%7D%7BV%28L%29%7D)
![[H^{+}] = \frac{0.761}{1.00}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B0.761%7D%7B1.00%7D)
![[H^{+}] = 0.761 \frac{mol}{L}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%200.761%20%5Cfrac%7Bmol%7D%7BL%7D)
From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows
![K_{w} = [H^{+} ][OH^{-} ]](https://tex.z-dn.net/?f=K_%7Bw%7D%20%3D%20%5BH%5E%7B%2B%7D%20%5D%5BOH%5E%7B-%7D%20%5D)
![[OH^{-}]=\frac{Kw}{[H^{+}] }](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D%5Cfrac%7BKw%7D%7B%5BH%5E%7B%2B%7D%5D%20%7D)
![[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D%5Cfrac%7B1.01X10-%5E%7B-14%7D%7D%7B0.761%20%7D)
![[OH^{-}]=1.33X10^{-14}\frac{mol}{L}](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D1.33X10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7BL%7D)
The pH of the solution can be measured by the following formula:
![pH = -log[H^{+} ]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%7B%2B%7D%20%5D)

