Answer:
Conductivity meter
Explanation:
A conductivity meter is normally used to measure the amount of electrical current or conductance in a solution. Conductivity is most useful in determining the overall health of a natural water body.
A pH paper is used to determine the pH of a solution. This is done by dipping part of the paper into a solution of interest and watching the color change. The pH paper comes in a color-coded scale indicating the pH that something has when the paper turns a certain color.
An indicator is an organic compound that changes its colour depending on the pH of the solution.
Since neutralization reaction can only be monitored by monitoring the pH of the solution, a conductivity meter cannot be used to monitor the progress of a neutralization reaction since it does not monitor the change in pH of the system under study.
Resources that come from nature, some examples include: Water, Gold, Oil, Coal, Apples, Oranges, etc.
Volume of Cl₂(g) is produced at 1.0 atm and 540.°C=4.5×10^4 L
As per the evenly distributed response
2NaCl (g) ----> 2Na(l)+ Cl2(g)
Calculate the amount of Cl2 that was formed as indicated below:
Moles of Cl2 = 31.0 kg of Na x (1000* 1 * 1 / 1*23* 2)
= 673.9 mol
P is equal to 1.0 atm, and T is equal to 813.15 K
when converted to Kelvin by multiplying by a factor of 273.15.
Using Cl2 as an ideal gas, determine the in the following volume:
volume = nRT/P
= 673.9 * 0.0821 * 813.15/ 1
=4.5×10^4 L
As a result, the volume of Cl2 under the given circumstances =4.5×10^4 L
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Answer:
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Explanation:
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3. 4 g of a nonelectrolyte dissolved in 78. 3 g of water produces a solution. The molar mass of the solute will be 17.94.
<h3>
What is molar mass?</h3>
Molar mass of a substance is its mass in grams in per mole of a solution.
Freezing point: Freezing point of a substance is a temperature at which a liquid starts to solidify.
Depression in the freezing point can be calculated
[Depression in freezing point of pure solvent—Freezing point of solution] =[(0) - (-4.5)] °C =4.5 °C
molar mass = Number of moles of solute m / Mass of solvent in Kg
3.4g / M x 1/ 0.0783 kg = 43.42
Substitute AT by 4.5°C , Kr by 1.86 °C/m, and m by 43.42 m in equation (1) as follows:
1.86 x 43.42 / 4.5 = 17.94
Therefore, molar mass of solute to be 17.94.
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