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chubhunter [2.5K]
3 years ago
5

Two objects are dropped into beaker with layers of oil, rubbing alcohol, water and corn syrup. These objects are made of the sam

e substance, but object X is twice as large, so twice as heavy as object Z.
Which of these two objects will settle further down into the beaker, X, Z, both or neither?
(My teacher labeled this as a trick question)
Chemistry
1 answer:
Mademuasel [1]3 years ago
6 0
It all depends on the objects' densities ,because if they're less dense than the oil they will not even sink to the bottom of the beaker
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Calculate the volume of a box with the following dimensions: length=8 cm
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Answer:  240 cm³

Explanation:

Volume of the box = l × w × h

                               = (8 cm × 6 cm × 5 cm)

                               = 240 cm³

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What is a common feature of elements in group 1, group 2 and group 3?​
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Answer:

all of them have seven valence electron

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What is the ground-state electron configuration of a neutral atom of manganese?.
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A 25.0 ml sample of 0.723 m hclo4 is titrated with a 0.27303 m koh solution. the h3o+ concentration after the addition of 66.2 m
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This doesn't need an ICE chart. Both will fully dissociate in water.

Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.

Step 1:

write out balanced equation for the reaction

HClO4+KOH ⇔ KClO4 + H2O

the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4

Step 2:

Determining the number of moles present in HClO4 and KOH

Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4

Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L

Remember:

M = moles/L so we have 0.025 L of 0.723 moles/L HClO4

Multiply the volume in L by the molar concentration to get:

0.025L x 0.723mol/L = 0.0181 moles HClO4.

Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH

Step 3:

Determine how much HClO4 remains after reacting with the KOH.

Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:

moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0

This means all of the HClO4 is used up in the reaction.

If all of the acid is fully reacted with the base, the pH will be neutral = 7.

Determine the H3O+ concentration:

pH = -log[H3O+]; [H3O+] = 10-pH = 10-7

The correct answer is 1.0x10-7.
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