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2 years ago

Which electron configuration is impossible?

Chemistry

1 answer:

Verizon [17]2 years ago

7 0

Answer:

MP hippopotamus

Explanation: I'm in the fifth grade so yeah I really don't know this so I'm just going to say some random stuff and white cheddar mac and cheese warm TV perfect back MD 11443 to 2885 eleven 12:20 to 11 132 0.24 answer

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A 10 gram sample of H20 is sealed in a 1350 ml flask at 27°C. Given the fact that water has a vapor pressure of 26.7 mmHg at thi

Aleksandr-060686 [28]

Answer:

9.9652g of water

Explanation:

The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:

n = PV / RT

Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)

Replacing:

n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K

n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:

1.93x10⁻³ moles × (18.01g / 1mol) = 0.0348g of water

As the initial mass of water was 10g, the mass of water that remains in liquid phase is:

10g - 0.0348g = 9.9652g of water

I hope it helps!

4 0

3 years ago

W(OH)2 + 2 HCl → WCl2 + 2 H2O

Afina-wow [57]

The amount of W(OH)2 needed would be 448.126 g

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

W(OH)2 + 2 HCl → WCl2 + 2 H2O

The mole ratio of W(OH)2 to HCl is 1:2

Mole of 150g HCl = 150/36.461

= 4.11 moles

Equivalent mole of W(OH)2 = 4.11/2

= 2.06 moles

Mass of 2.06 moles W(OH)2 = 2.06 x 217.855

= 448.188g

More on stoichiometric calculations can be found here: brainly.com/question/8062886

7 0

2 years ago

2(x-1 ➗ x+3)-7(x+3 ➗ x-1)=5

NeTakaya

I attached a photo of my work below. Please tell me if I got it wrong; hope you got a good score on your assignment! ^v^

3 0

3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

How is this model useful?

const2013 [10]

Answer:

A) It shows how electrons are distributed in the Shells of an Iron atom.

Explanation: Took the test on Edge

6 0

3 years ago