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3 years ago

What is the empirical formula of the compound whose molecular formula is P4O10

Chemistry

2 answers:

Pepsi [2]3 years ago

7 0

The empirical formula is the lowest reduced ration of the atoms present. In this case, it is P2O5

9966 [12]3 years ago

6 0

Answer:

The empirical formula of P4O10 is P2O5 . This is arrived at by dividing both subscripts by their highest common factor 2.

Explanation:

Empirical Formula of a compound is the lowest whole number ratio between the compound elements it is sometimes different from the standard formula of the compound).

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Please somebody give me the answers

RSB [31]

Answer:

1. 9.4 grams of methane produce 25.85 grams of CO2

2.Grams of water produced = 11.81 grams

3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams

4.Amount of methane consumed = 46.9 grams

5. Grams of Co2 produced = 8.32 grams

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

1 mole of CH4 = 16 gram

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams (1 mole of CO2 = 44 gram )

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

$1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)$

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

$\frac{44}{16}$ gram of CO2

9.4 gram of CH4 =

$\frac{44}{16}\times 9.4$ gram of CO2

= 25 .85 gram of CO2

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

$\frac{18}{32}$

21 gram of O2 =

$\frac{18}{32}\times 21$

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32) = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

$\frac{16}{64}$

10.1 gram of O =

$\frac{16}{64}\times 10.1$ of CH4

= 2.52 gram

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

$\frac{16}{44}$ gram of CH4

129 gram of CO2 =

$\frac{16}{44}\times 129$ gram of CH4

= 46.90 grams

2 mole of O2 produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

$\frac{44}{64}$ of CO2

12.1 gram of O2 produce=

$\frac{44}{64}\times 12.1$ of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

8 0

3 years ago

How would you test a colorless crystalline compound to determine if it was a hydrate?

Cloud [144]

A hydrate is a substance where in it contains water and other constituent elements. To know whether if that compound was a hydrate,you should record its mass, then put it in a test tube and heat it with a Bunsen burner. If the compound is a hydrate, the water in the compound will discharge in the form of water vapor. At the next 5-10 minutes, remove it in the test tube and weigh it up again. If the mass is now fewer, that means that there was water existing that has now evaporated, and the compound was a hydrate.

6 0

3 years ago

What two elements in the sun mostly made of

ivanzaharov [21]

The sun is made up of two main elements, hydrogen and helium.

Hydrogen makes up about 92% of all of the atoms in the sun while helium makes up about 7.8%.
Oxygen, carbon, neon and nitrogen make up most of the remaining 0.2%.

8 0

3 years ago

3. A student measured 15.0 grams of ice in a beaker. The beaker was then

gregori [183]

Answer:

The heat that was used to melt the 15.0 grams of ice at 0°C is 4,950 Joules

Explanation:

The mass of ice in the beaker = 15.0 grams

The initial temperature of the ice = 0°C

The final temperature of the ice = 0°C

The latent heat of fusion of ice = 330 J/g

The heat required to melt a given mass of ice = The mass of the ice to be melted × The latent heat of fusion of ice

Therefore, the heat, Q, required to melt 15.0 g of ice = 15.0 g × 330 J/g = 4,950 J

The heat that was used to melt the 15.0 grams of ice = 4,950 Joules.

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3 years ago

Im sooooooooooooooooooooooooo cufused on science

Elena-2011 [213]

Answer:

What do you need please to understand?

6 0

3 years ago