Answer:
C.)
Explanation:
The general equilibrium expression looks like this:
In this expression,
-----> K = equilibrium constant
-----> uppercase letters = formulas
-----> lowercase letters = balanced equation coefficients
Solids are not included in the expression, so Ca(HCO₃)₂(s) and CaO(s) should be left out. The products should be in the numerator and the reactants are in the denominator. This makes the expression:
<----- Equilibrium expression
<----- Insert gaseous formulas
<----- Insert coefficients
It had to be a physical change unless there a chemical touching the metal surface making it a chemical change .
Answer:
In a controlled experiment, an independent variable (the cause) is systematically manipulated and the dependent variable (the effect) is measured; any extraneous variables are controlled. State what steps you will take to design an experiment to test your hypothesis. Complete the experiment. Make observations. Record what happened during your experiment.
The volume of the liquid in each cylinder are given below
A. 48.9 mL
B. 48.5 mL
C. 11 mL
<h3>Shape of liquid in a cylinder</h3>
A liquid in a cylinder tends to show a curved shape due to adhesion.
Hence, to get an accurate measurement, we must take the reading at the lower meniscus (i.e under the curve shape of the liquid)
<h3>A. How to determine the volume of the cylinder</h3>
From the diagram given, we can see that the lower meniscus (i.e under the curve) is 48.9 mL since 10 lines exist between 48 and 49. Thus, the volume of the liquid is 48.9 mL
<h3>B. How to determine the volume of the cylinder</h3>
From the diagram given, we can see that the lower meniscus (i.e under the curve) is 48.5 mL since 10 lines exist between 48 and 49. Thus, the volume of the liquid is 48.5 mL
<h3>C. How to determine the volume of the cylinder</h3>
From the diagram given, we can see that the lower meniscus (i.e under the curve) is 11 mL since 5 lines exist between 10 and 15. Thus, the volume of the liquid is 11 mL
Learn more about reading volume of liquid:
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The molarity of 25.00 grams of powdered AgNO3 added to 500.00mL of pure water is 0.29 mol/L
<h3>Stoichiometry</h3>
From the question, we are to determine the molarity of the solution prepared.
First, we will determine the number of moles of AgNO₃ present.
Using the formula,
From the given information,
Mass = 25.00 g
Molar mass = 169.874 g/mole
Number of moles =
Number of moles of AgNO₃ present = 0.147168 mole
Now, for the molarity of the solution
From the given information,
Volume = 500.00 mL = 0.5 L
Molarity =
Molarity = 0.294336
Molarity ≅ 0.29 mol/L
Hence, the molarity of 25.00 grams of powdered AgNO3 added to 500.00mL of pure water is 0.29 mol/L
Learn more on Stoichiometry here: brainly.com/question/14805986