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3 years ago

Which of the following terms describes a piece of pure gold?

Chemistry

2 answers:

vladimir2022 [97]3 years ago

6 0

Answer:

The correct answer is the option: d. mixture.

Explanation:

Hello! Let's solve this!

Pure gold is a mixture, since it is mixed with some other metal to increase its characteristics.

Pure gold or fine gold is 24-karat gold.

A mixture is the union of one or more elements, but they are not chemically linked.

We conclude that the correct answer is the option: d. mixture.

SCORPION-xisa [38]3 years ago

4 0

Mixture would be the piece of pure gold. I don't know the explanation but the answer is mixture, I'm sure about that it is mixture, I read it somewhere.

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3 years ago

8.310x10^2 – 7.210x10^1<br><br>[?]x10^[?]

alisha [4.7K]

Answer:

$7.589\times 10^{2}$

Explanation:

The expression can be solved mathematically as follows:

1) $8.310\times 10^{2}-7.210\times 10^{1}$ Given

2) $8.310\times 10^{1+1} - 7.210\times 10^{1}$ Definition of sum

3) $(8.310\times 10^{1})\times 10^{1}-7.210\times 10^{1}$ $a^{m+n} = a^{m}\cdot a^{n}$ /Associative property

4) $(8.310\times 10^{1}-7.210)\times 10^{1}$ Distributive property

5) $(83.100-7.210)\times 10^{1}$ Multiplication

6) $75.89\times 10^{1}$ Subtraction.

7) $(7.589\times 10^{1})\times 10^{1}$ Multiplication/Associative property

8) $7.589\times (10^{1}\times 10^{1})$ Associative property

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3 years ago

How did Thomson discover the electron?

Hoochie [10]

J. J. Thomson discovered electron by performing an experiment using cathode ray tubes. High voltage across is applied across two electrodes at that causes a beam of particles to flow from the the negatively-charged electrode that is cathode to the positively-charged electrode that is anode. Properties of the particles, are tested using two oppositely-charged electric plates around the cathode ray. The cathode ray was deflected away from the negatively-charged electric plate and towards the positively-charged plate. This indicated that the cathode ray was composed of negatively-charged particles. And these negatively charged particles are called electrons.

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4 years ago

A sample of a compound contains 3.21 g of sulfur and 11.4 g of fluorine. Which of the following represents the empirical formula

Sergeeva-Olga [200]

Answer:

The empirical formula is SF6 (option E)

Explanation:

Step 1: Data given

Mass of sulfur = 3.21 grams

Mass of fluorine = 11.4 grams

Molar mass sulfur = 32.065 g/mol

Molar mass fluorine = 19.00 g/mol

Step 2: Calculate moles

Moles = mass /molar mass

Moles sulfur = 3.21 grams / 32.065 g/mol

Moles sulfur = 0.100 moles

Moles fluorine = 11.4 grams / 19.00 g/mol

Moles fluorine = 0.600 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.100 / 0.100 = 1

F : 0.600 / 0.100 = 6

The empirical formula is SF6 (option E)

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3 years ago

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