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4 years ago

Ozone(O3) number of atoms

Chemistry

1 answer:

spayn [35]4 years ago

8 0

As the O3 suggests, the Ozone molecule is made up of 3 Oxygen atoms.

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Why is it important for scientists to describe their procedures in full detail?

Licemer1 [7]

So that other scientist can repeat their experiments
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3 years ago

What’s potassium+calcium sulphate

Dvinal [7]

Potassium +calcium sulphate ------> K2SO4+Ca

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3 years ago

Which pair of elements have the same number of atomic orbitals?

levacccp [35]

Answer:

Sodium and strontium

Explanation:

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3 years ago

A sample of neon is collected at 2.7 atm and 12.0 oC. It has a volume of 2.25 L. What would be the volume of this gas at STP?

Keith_Richards [23]

Answer:

5.82 L

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 2.25 L

V₂ = ?

P₁ = 2.7 atm

T₁ = 12 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (12 + 273.15) K = 285.15 K

At STP, Pressure = 1 atm and Temperature = 273.15 K

So,

P₂ = 1 atm

T₂ = 273.15 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{2.7}\times {2.25}}{285.15}=\frac {{1}\times {V_2}}{273.15}$

Solving for V₂ , we get:

7 0

4 years ago

Groundwater in Pherric, New Mexico, initially contains 1.800 mg/L of iron as Fe3+. What must the pH be raised to in order to pre

xenn [34]

Answer : The pH will be, 3.2

Explanation :

As we known that the value of solubility constant of ferric hydroxide at $25^oC$ is, $2.79\times 10^{-39}$

Amount or solubility of iron consumed = (1.800 - 0.3) mg/L = 1.5 mg/L

The given solubility of iron convert from mg/L to mol/L.

$1.5mg/L=\frac{1.5\times 10^{-3}g/L}{56g/mol}=2.7\times 10^{-7}mol/L$

The chemical reaction will be:

$Fe(OH)_3\rightarrow Fe^{3+}+3OH^-$

The expression of solubility constant will be:

$K_{sp}=[Fe^{3+}]\times [3OH^-]^3$

Now put all the given values in this expression, we get the concentration of hydroxide ion.

$2.79\times 10^{-39}=(2.7\times 10^{-7})\times [3OH^-]^3$

$[OH^-]=1.5\times 10^{-11}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (1.5\times 10^{-11})$

$pOH=10.8$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-10.8\\\\pH=3.2$

Therefore, the pH will be, 3.2

4 0

4 years ago