In a physical change the appearance or form of the matter changes but the kind of matter in the substance does not. However in a chemical change, the kind of matter changes and at least one new substance with new properties is formed. Hope this helps!
Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
Nonmetals which are located in the second row form pi bonds
more easily than the elements situated in the third row and below. Actually there
are no compounds or molecules known that forms covalent bonds to the noble gas
Ne and Ar. Hence the other second row element which is Carbon, is the element that
forms
pi bonds most readily.
Answer:
<span>C</span>
Answer:
A. How much decay happens in each flashlight
Explanation:
In an experiment, the DEPENDENT OR RESPONDING VARIABLE is the variable that responds to changes being made to another variable called Independent variable. It is the variable that is measured by an experimenter.
In this question, an experiment was carried out to research the effect of sunlight on plastic decay. Three plastic flashlights were used for this experiment in which one was placed in a dark closet (control), and the others were placed in light at different times or intervals. However, the variable being measured or assessed is the decay of each flashlight. This means that "HOW MUCH DECAY HAPPENS IN EACH FLASHLIGHT" is the dependent variable.
<h3><em>Hope</em><em> it</em><em> helps</em><em> you</em><em>.</em></h3>
<em> </em><em> </em><em> </em><em> </em><em>do </em><em>follow</em><em> me</em><em> </em><em>if </em><em>you </em><em>appreciate</em><em> it</em><em> </em>