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3 years ago

How many orbitals does fluorine contain

Chemistry

2 answers:

Nata [24]3 years ago

7 0

Answer:

Explanation:

Lisa [10]3 years ago

6 0

Hi there!

We have three 2p orbitals in fluorine.

Hope this helps !

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Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +

nikklg [1K]

Answer:

For 1: The value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

For 2: The value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

Explanation:

For the given chemical equation:

$H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)$

For 1:

To calculate the $\Delta G$ for given value of equilibrium constant, we use the relation:

$\Delta G=-RT\ln K_p$ .....(1)

where,

$\Delta G$ = ? kJ/mol

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

For 2:

The expression of $K_p$ for the given chemical equation is:

$K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}$

We are given:

$p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm$

Putting values in above equation, we get:

$K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52$

Now, calculating the value of $\Delta G$ by using equation 1:

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

3 0

3 years ago

Read 2 more answers

You are performing an experiment that uses 114Ag. 113Ag is radioactive, decays by beta-- emission and has a half-life of 21 minu

maxonik [38]

Answer: The 234.74 grams of sample should be ordered.

Explanation:

Let the gram of 114 Ag to ordered be $N_o$

The amount required for the beginning of experiment = 0.0575 g

Time requires to ship the sample = 4.2hour = 252 min(1 hr = 60 min)

Half life of the sample = ${t_{\frac{1}{2}}$ = 21 min

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{21 min}=0.033 min^{-1}$

$\log[N]=\log[N_o]-\frac{\lambda t}{2.303}$

$\log[0.0575 g]=\log[N_o]-\frac{0.033 min^{-1}\times 252 min}{2.303}$

$N_o=234.74 grams$

The 234.74 grams of sample should be ordered.

4 0

3 years ago

The chemical equation below shows the formation of aluminum oxide (Al2O3) from aluminum (Al) and oxygen (O2).

Rufina [12.5K]

Answer:

Mass = 182.4 g

Explanation:

Given data:

Number of moles of Al₂O₃ = 3.80 mol

Mass of oxygen required = ?

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Now we will compare the moles of aluminum oxide and oxygen.

Al₂O₃ : O₂

2 : 3

3.80 : 3/2×3.80 = 5.7

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 5.7 mol × 32 g/mol

Mass = 182.4 g

4 0

2 years ago

What are the safety procedures for nuclear accidents nowadays?

Zigmanuir [339]

Explanation:

Take shelter in a hard wall building

Close doors and windows cut off ventilation

4 0

3 years ago

Independent variable

katrin [286]

Answer:

Independent variable

aluminium ball

6 0

3 years ago