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kirill115 [55]
3 years ago
11

Please help me with these two questions

Chemistry
1 answer:
ANEK [815]3 years ago
8 0
What is the question?
You might be interested in
Electron configuration for Bohr model for sodium is
MrMuchimi

Answer: The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the remaining electron in the 3s. Therefore the sodium electron configuration will be 1s22s22p63s1.

Explanation:

3 0
3 years ago
What is the electron configuration for oxygen with a 2- charge (o2-)?
Juli2301 [7.4K]
Oxygen:
Atomic no. = 8(from periodic table)
⇒1s^2 2s^2 2p^4
But it is O^2-
There are 2 more electrons
=>1s^2 2s^2 2p^6
Voila!
3 0
3 years ago
A gas occupies 100 mL at 150. kPa. Find its volume at 200. kPa. You must show all your work to receive credit. Be sure to identi
PtichkaEL [24]

The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:

PV = k

where k is a constant.

We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,

P₁V₁ = P₂V₂

Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.

In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa.  Substituting these values in the previous equation and clearing V₂, we have,

P₁V₁ = P₂V₂ → V₂ = \frac{P1V1}{P2}  

→ V₂ = \frac{100 mL x 150 kPa}{200 kPa}

→ V₂ = 75 mL

Then, the volume occupied by the gas at 200 kPa is V₂ = 75 mL

6 0
3 years ago
What two substances are formed at the end point of a neutralization reaction
Slav-nsk [51]
Most likely the answer is water and a salt. When a strong acid and strong base react the OH and H form water and the cation from the base and anion from the acid forms a salt. I hope this helps. Let me know if anything is unclear.
6 0
3 years ago
When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
3 years ago
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