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omeli [17]
3 years ago
6

6.002cm= what to sig figs

Chemistry
1 answer:
kirza4 [7]3 years ago
7 0
6.002 is equals to 4 significant figures.
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Freezing a candy bar is an example of a physical change. <br> True or false
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Answer: This is true.

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 It is true because if becomes frozen, then it is physically harder to melt...

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2 years ago
How many particles are in 3.2 mole of neon gas
Cerrena [4.2K]

There are 1.93 x 10²⁴ particles

<h3>Further explanation</h3>

Given

3.2 moles of Neon gas

Required

Number of particles

Solution

The mole is the number of particles(molecules, atoms, ions) contained in a substance  

<em>1 mol = 6.02.10²³ particles </em>

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

So the number of particles for 3.2 moles :

N = 3.2 x 6.02.10²³

N = 1.93 x 10²⁴

or

we can describe it using Avogadro's number conversion factor

\tt 3.2~moles\times \dfrac{6.02\times 10^{23}}{1~mole}=1.93\times 10^{24}

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3 years ago
14. What is a major disadvantage of using solar power?
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Methyl salicylate is a common active ingredient in liniments such as ben-gay. It is also known as oil of wintergreen. It is made
arlik [135]

Answer: The empirical formula is C_3H_3O.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2 = 12.24 g

Mass of H_2O = 2.505 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 12.24 g of carbon dioxide, =\frac{12}{44}\times 12.24=3.338g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.505 g of water, =\frac{2}{18}\times 2.505=0.278g of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671  g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.278}{0.104}=3

For H =\frac{0.278}{0.104}=3

For O =\frac{0.104}{0.104}=1

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is C_3H_3O.

5 0
3 years ago
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