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3 years ago

How many atoms are present in a Neon molecule? *

Chemistry

1 answer:

Ivan3 years ago

4 0

Answer:

Neon molecule does not exist.

Explanation:

Neon is denoted by Ne which is noble gas. According to the "Molecular Orbital Theory", only those molecules exist that have a positive bond order and molecules with no bond or negative bond order cannot exist.

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Calculate the percentage composition of Mg3 (Po4)2

lutik1710 [3]

Mg3(PO4)2 - the molar mass would be 262g/mol, which is 100%

Atomic mass of Mg is 24, since we have 3Mg we multiply by 3 and get a mass of 72

262 : 100% = 72 : x%

x = 72*100 / 262

x = 27.5%

And do that for every element — get the molar mass of P and multiply by 2, use a ratio, and get the molar mass of O and multiply by 8 and use ratios :)

7 0

3 years ago

What is climate? How does it differ from weather

ozzi

Weather refers to short term atmospheric conditions while climate is the weather of a specific region averaged over a long period of time. Climate change refers to long-term changes.

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3 years ago

Read 2 more answers

A gas has a volume of 490. mL at a temperature of -35.0 degrees C. What volume would the gas occupy at 42.0 degrees Celsius? Ple

miskamm [114]

Answer:

648.5 mL

Explanation:

Here we will assume that the pressure of the gas is constant, since it is not given or specified.

Therefore, we can use Charle's law, which states that:

"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"

Mathematically:

$\frac{V}{T}=const.$

where

V is the volume of the gas

T is its absolute temperature

The equation can be rewritten as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where in this problem we have:

$V_1=490 mL$ is the initial volume of the gas

$T_1=-35.0^{\circ} + 273 = 238 K$ is the initial temperature

$T_2=42.0^{\circ}+273=315 K$ is the final temperature

Solving for V2, we find the final volume of the gas:

$V_2=\frac{V_1 T_2}{T_1}=\frac{(490)(315)}{238}=648.5 mL$

8 0

4 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

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 $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

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 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

56789x09876 Does anyone want to rp?? Any kind of rp I'm just bored

Alja [10]

The answer would be 560,848,164 because the 0 at the start WOULDNT matter

7 0

3 years ago