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katen-ka-za [31]
3 years ago
7

Help me answer this chemistry question. thank you!

Chemistry
1 answer:
Leona [35]3 years ago
7 0

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.

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B. double-replacement RXN

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more specifically, this is a precipitation rxn.

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Is neon a metal, nonmetal , or metalloid
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what is the concentration of hydroxide ions after 50.0 ml of 0.250 m naoh is added to 120 ml of 0.200 m na2so4? please show all
Galina-37 [17]

The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

What is meant by concentration?

Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.

Concentration of hydroxide ions can be calculated by,

M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.

where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.

Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

To learn more about concentration click on the given link brainly.com/question/17206790

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exo = exit
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Sorry if I'm wrong, hope I helped. :)
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