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aleksklad [387]
3 years ago
6

a classmate examines a substance that conducts electricity and is shiny but has no color. He says that is must be silver. What i

s his error?
Chemistry
2 answers:
saul85 [17]3 years ago
6 0

Answer:

Not all materils that conduct electricity are silver.

Explanation:

The conductive materials are many and different. The best known are metals such as silver, gold, copper, aluminum.

But some aqueous solutions of salts are also conductive, even the human body is conductive.

As you can imagine, not all of these materials or solutions are silver or even shiny. So the mistake is in concluding that all conductive material must be silver and shiny, when that is not always the case.

Dovator [93]3 years ago
5 0

He is making a conclusion from a single observation.

Silver isn’t the only metal that is shiny and has no colour.

For example, lithium, sodium, magnesium, and aluminium all have about the same colour as silver.

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How many grams of oxygen are needed to react with 4.6 grams of titanium(IV) chloride
velikii [3]
The answer is: 0.78g
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3 years ago
Rn-222 has a half-life of 3.82 days. If 25.0 g of Radon-22 was originally present, approximately how many grams would be left af
horsena [70]

Answer:

Approximately 0.39 g or 0.4 g if you're rounding up

Explanation:

15/3.82 = 3.92

Let's round that up to 4

That means 15 days is around 4 half lives

4 half lives means 1/16 of the original mass will be left

25/16 = 0.390625

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4 years ago
Is this molecule polar or nonpolar?
lutik1710 [3]
The answer is nonpolar  

3 0
3 years ago
Which of the following sets of quantum numbers is NOT allowed? a. n = 5, l= 4, ml= –2, ms = +1/2 b. n = 2, l = 1, ml= 0, ms = +1
klemol [59]

Answer:

C

Explanation:

the n value must always be greater than the l value

8 0
4 years ago
How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)
zhannawk [14.2K]

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0
3 years ago
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