Answer:
He will decide which drink is to be served to whom, by the use of litmus paper.
Explanation:
The litmus paper is the most common indicator to determine the acidity or basicity of a solution. Blue litmus paper changes its color to red when a solution changes from basic to acidic while red litmus paper changes its color to blue when the opposite occurs (acid → basic).
First of all the litmus paper strip, pH indicator, is immersed in a solution and allowed to pass between 10 and 15 seconds while keeping the strip submerged. Afterwards it is removed, and then the strip compares the color. If the color is diffuse, there is a color scale where it is determined which solution has alkaline or acidic pH
Answer:
The Solar System moves through the galaxy with about a 60° angle between the galactic plane and the planetary orbital plane. The Sun appears to move up-and-down and in-and-out with respect to the rest of the galaxy as it revolves around the Milky Way
Explanation:
Hope you like it
Answer:
<h3>The answer is 7.42 </h3>
Explanation:
The pH of a solution can be found by using the formula
![pH = - log [ { H_3O}^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5B%20%7B%20H_3O%7D%5E%7B%2B%7D%5D)
From the question we have
![pH = - log(3.8 \times {10}^{ - 8} ) \\ = 7.420216...](https://tex.z-dn.net/?f=pH%20%3D%20%20-%20%20log%283.8%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%208%7D%20%29%20%20%5C%5C%20%20%3D%207.420216...)
We have the final answer as
<h3>7.42 </h3>
Hope this helps you
Answer:
![\large \boxed{\text{B.) 2.8 atm}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7BB.%29%202.8%20atm%7D%7D)
Explanation:
The volume and amount are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.
![\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}](https://tex.z-dn.net/?f=%5Cdfrac%7Bp_%7B1%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cdfrac%7Bp_%7B2%7D%7D%7BT_%7B2%7D%7D)
Data:
p₁ = 1520 Torr; T₁ = 27 °C
p₂ = ?; T₂ = 150 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 27 + 273.15) K = 300.15 K
T₂ = (150 + 273.15) K = 423.15 K
(b) Calculate the new pressure
![\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cdfrac%7B1520%7D%7B300.15%7D%20%26%20%3D%20%26%20%5Cdfrac%7Bp_%7B2%7D%7D%7B423.15%7D%5C%5C%5C%5C5.064%20%26%20%3D%20%26%20%5Cdfrac%7Bp_%7B2%7D%7D%7B423.15%7D%5C%5C%5C%5C5.064%5Ctimes423.15%26%3D%26p_%7B2%7D%5C%5Cp_%7B2%7D%20%26%20%3D%20%26%20%5Ctext%7B2143%20Torr%7D%5Cend%7Barray%7D%5C%5C)
(c) Convert the pressure to atmospheres
![p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}](https://tex.z-dn.net/?f=p%20%3D%20%5Ctext%7B2143%20Torr%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20atm%7D%7D%7B%5Ctext%7B760%20Torr%7D%7D%20%3D%20%5Ctextbf%7B2.8%20atm%7D%5C%5C%5C%5C%5Ctext%7BThe%20new%20pressure%20reading%20will%20be%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B2.8%20atm%7D%7D%24%7D)