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Serhud [2]
3 years ago
6

Plz help me with this:

Chemistry
1 answer:
In-s [12.5K]3 years ago
7 0

Answer:

1) H 1s^1

2) [Ar] 3d⁸ 4s²

3) [He] 2s2 2p3

4) [Kr] 4d10 5s2 5p5

5) [Ar] 4s² and [Ar] 4s²

6) [He] 2s2 2p2

7)  [He] 2s² 2p⁴

8) [Ar] 3d7 4s2

9) [Kr] 4d¹⁰ 5s¹  

10) [Ne] 3s² 3p⁶

Explanation:

You might be interested in
Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).
Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas         Mole Fraction      kH mol/(L*atm)

           N_2                         7.81\times 10^{-1}         6.70\times 10^{-4}

           O_2                         2.10\times 10^{-1}        1.30\times 10^{-3}

           Ar                          9.34\times 10^{-3}        1.40\times 10^{-3}

          CO_2                        3.33\times 10^{-4}        3.50\times 10^{-2}

          CH_4                       2.00\times 10^{-6}         1.40\times 10^{-3}

          H_2                          5.00\times 10^{-7}         7.80\times 10^{-4}

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}

where,

p_A = partial pressure of hydrogen gas = ?

p_T = total pressure = 0.380 atm

\chi_A = mole fraction of hydrogen gas = 5.00\times 10^{-7}

Putting values in above equation, we get:

p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{H_2}=K_H\times p_{H_2}

where,

K_H = Henry's constant = 7.80\times 10^{-4}mol/L.atm

p_{H_2} = partial pressure of hydrogen gas = 1.9\times 10^{-7}atm

Putting values in above equation, we get:

C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

4 0
3 years ago
Many enzymes are inhibited irreversibly by heavy metal ions such as Hg2+ , Cu2+ , or Ag+ , which can react with essential sulfhy
3241004551 [841]

Answer:

The minimum molecular weight of the enzyme is 29.82 g/mol

Explanation:

<u>Step 1:</u> Given data

The volume of the solution = 10 ml = 10*10^-3L

Molarity of the solution = 1.3 mg/ml

moles of AgNO3 added = 0.436 µmol = 0.436 * 10^-3 mmol

<u>Step 2:</u> Calculate the mass

Density = mass/ volume

1.3mg/mL = mass/ 10.0 mL

mass = 1.3mg/mL *10.0 mL = 13mg

<u>Step 3:</u> Calculate minimum molecular weight

Molecular weight = mass of the enzyme / number of moles

Molecular weight of the enzyme = 13mg/ 0.436 * 10^-3 mmol

Molecular weight = 29.82 g/mole

The minimum molecular weight of the enzyme is 29.82 g/mol

7 0
3 years ago
13. What physical property is characteristic of all of the elements in the group located in the
zimovet [89]

Answer:

B. gas state at room temperature

Explanation:

6 0
2 years ago
2. A quantity of 1.922g of methanol (CH3OH) was burned in a constant-volume
Cerrena [4.2K]
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
5 0
2 years ago
A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many m
oksian1 [2.3K]

Answer : The 866.66\mu L must be administered.

Solution :

As we are given that a vial containing radioactive selenium-75 has an activity of 3.0mCi/mL.

As, 3.0 mCi radioactive selenium-75 present in 1 ml

So, 2.6 mCi radioactive selenium-75 present in \frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L

Conversion :

(1ml=1000\mu L)

Therefore, the 866.66\mu L must be administered.

4 0
2 years ago
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