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2 years ago

1. Oxygen full electron configuration

Chemistry

1 answer:

Arada [10]2 years ago

7 0

Answer:

$1s^{2} 2s^{2} 2p^{4}$

Explanation:

Oxygen has eight eletrons and six valence electrons, giving it the electron configuration of $1s^{2} 2s^{2} 2p^{4}$ .

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EXTRA CREDIT

marin [14]

Answer:

$\huge\boxed{\sf 36\ H\ atoms}$

Explanation:

Molecular formula from Glucose:

C₆H₁₂O₆

3 moles of Glucose:

3C₆H₁₂O₆

In 1 mole of Glucose, there are 12 hydrogen atoms.

In 3 moles:

= 12 × 3

= 36 H atoms

$\rule[225]{225}{2}$

5 0

2 years ago

Determine the [H3O+] in a 0.265 M HClO solution. The Ka of HClO is 2.9 × 10-8.

dexar [7]

8.8 × 10-5 M is the [H3O+] concentration in 0.265 M HClO solution.

Explanation:

HClO is a weak acid and does not completely dissociate in water as ions.

the equation of dissociation can be written and ice table to be formed.

HClO +H2O ⇒ ClO- + H3O+

I 0.265 0 0

C -x +x +x

E 0.265-x +x +x

Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.

Ka = $\frac{[ClO-][H3O+]}{[HClO]}$

2.9 × 10^-8 = $\frac{[x] [x]}{[0.265-x]}$

$x^{2}$ = 7.698 x $10^{-9}$

x = 8.8 × 10-5 M

The hydronium ion concentration is 8.8 × 10-5 M in 0.265 M solution of HClO.

8 0

2 years ago

What is a scientific question?

Lunna [17]

Answer:

im guessing it's the second one

5 0

3 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

Answer: The freezing point of solution is -0.454°C

Explanation:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

3 0

2 years ago

Examine the chemical reaction and lab scenario.

Ierofanga [76]

Explanation:

Percentage Yield

= (3.37g/3.81g) * 100% = 88.45%.

Therefore 88.45% SO2 is the percentage yield.

3 0

3 years ago