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olchik [2.2K]
3 years ago
13

The element Oxygen, O, is an example of which of the following?

Chemistry
2 answers:
Lady_Fox [76]3 years ago
6 0
It is an example of a molecule. It is a diatom meaning it is expressed as 0^2
Anon25 [30]3 years ago
5 0
It is an example of a molecule 
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A chemical equation that does not indicate relative amounts of reactants and products is called
Zielflug [23.3K]

Answer:

The correct answer is skeleton equation.

Explanation:

In chemistry, the skeletal formula of a compound is an abbreviated representation of its molecular structure. Skeleton formulas are used because they clearly show complicated structures, they are fast and simple to draw.

All atoms that are not carbon or hydrogen are represented by their chemical symbol. The relative amounts of reagents and products are not indicated.

Have a nice day!

6 0
3 years ago
What is the most likely metal ion present in the fireworks shown in the image?
Tasya [4]
A because of the way it looks and how it works
8 0
3 years ago
Read 2 more answers
A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o
Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0
3 years ago
How many kilograms of a fertilizer are made of pure P2O5 would be required to supply 1.69 kilogram of phosphorus to the soil?
Alekssandra [29.7K]

Answer:

3.89 kg P2O5 must be used to supply 1.69 kg Phosphorus to the soil.

Explanation:

The molecular mass of P2O5 is

P2 = 2* 31 =           62

O5 = 5 *<u> 16 =         80</u>

Molecular Mass = 142

Set up a Proportion

142 grams P2O5 supplies 62 grams of phosphorus

x    kg P2O5        supplies 1.69 kg of phosphorus

Though this might be a bit anti intuitive, you don't have to convert the units for this question. The ratio is all that is important.

142/x = 62/1.69            Cross multiply

142 * 1.69 = 62x           combine the left

239.98 = 62x               Divide by 62

239.98/62 = x

3.89 kg of P2O5 must be used.

3 0
3 years ago
Please help!
Dahasolnce [82]

Answer:

reproduction

Explanation:

reproduction, process by which organisms replicate themselves

6 0
2 years ago
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