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Neporo4naja [7]
3 years ago
10

n the table below, write the density of each object. Then predict whether the object will float or sink in each of the fluids. W

rite “Float” or “Sink” in each empty box of the table. Object Object density Liquid density 0.5 g/mL 1.0 g/mL 2.0 g/mL 1 1.4 sink sink float 2 3 0.8 sink float float 4 5
Chemistry
1 answer:
Scrat [10]3 years ago
3 0

Answer:

0.5 g/mL----- float

1.0 g/mL---- float

2.0 g/mL-----sink

Explanation:

Those objects will float whose density is less or equal to water density due to lower mass while those object will sink whose density is greater than water because the mass of the object is greater than water. So the density of the object i. e. 0.5 g/mL and 1.0 g/mL will float in the liquid because its density is lower than water which is 1 g/mL while the density of an object i. e. 2.0 g/mL is denser than water so it will sink.

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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
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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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