Lewis Structure of HOClO
Answers
Next thing that you should do is to connect the dots of HOClO and do not forget that each dot connect with one lone electron.
Then draw valence of each electron and connect them with a line.
Final answer:
The Lewis structure of HOClO consists of a hydrogen atom bonded to an oxygen atom, which is in turn bonded to a chlorine atom. Two chlorine atoms are also bonded to each other in the structure.
Explanation:
The Lewis structure of HOClO can be determined using the concept of valence electrons. Oxygen (O) has 6 valence electrons, Chlorine (Cl) has 7 valence electrons, and Hydrogen (H) has 1 valence electron. Based on this, the Lewis structure of HOClO can be represented as follows:
H - O - Cl - O
Where the oxygen atom is bonded to the hydrogen atom and the chlorine atom, and the two chlorine atoms are bonded to each other, resulting in a single bond and two double bonds.
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Which of the following metal atoms would have the highest conductivity?A.the atom with the smallest radius and the fewest valence electrons B.the atom with the smallest radius and the most valence electrons C.the atom with the largest radius and the fewest valence electrons D.the atom with the largest radius and the most valence electrons
Answers
(NH₄)₃PO₄ = 149.0 g/mol
Ca₃(PO₄)₂ = 310.17 g/mol
2 (NH₄)₃PO4 + 3 Ca(OH)₂= Ca₃(PO₄)₂ + 6 NH₄OH
2 * 149 g --------- 310.17 g
1.5 g -------------- ?
mass = 1.5 * 310.17 / 2 * 149
mass = 465.255 / 298
= 1.561 g of Ca₃(PO₄)₂
hope this helps!
(2) 50 K and 600 kPa (4) 750 K and 600 kPa
Answers
Answer : The correct option is, (3) 750 K and 20 kPa
Explanation :
The conditions for ideal gas are :
Ideal gas are those gas that has no intermolecular attractions.
Ideal gas are those gas that have negligible volume.
The ideal gas equation is,
The conditions for real gas are :
Real gas are those gas that have intermolecular attractions.
Real gas are those gas that have volume.
The real gas equation is,
A real gas behave ideally at high temperature and low pressure condition.
From the given options, option (3) have high temperature and low pressure is the correct option.
Hence, at 750 K and 20 kPa conditions of temperature and pressure does a sample of helium behave most like an ideal gas.
Answers
oxygen is the answer
B) The number of molecules decreases.
C) The temperature decreases.
D) The volume decreases.
Answers
Answer:
D)
Explanation:
The volume decreases.
Answers
Answer:
Explanation:
1 mol of helium contains 6.022 × 10²³ atoms
The balloon contains .
69.253 helium particles are in the balloon.
What is helium particles?
Helium is composed of two electrons bound by the electromagnetic force to a nucleus containing two protons along with either one or two neutrons, depending on the isotope, held together by the strong force. Unlike for hydrogen, a closed-form solution to the Schrodinger equation for the helium atom has not been found.
A helium atom is an atom of the chemical element helium. Helium is composed of two electrons bound by the electromagnetic force to a nucleus containing two protons along with either one or two neutrons, depending on the isotope, held together by the strong force. Unlike for hydrogen, a closed-form solution.
Helium is the element which you can find on the upper right side of the periodic table with atomic number 2. It comes first amongst the family of the noble gases. Helium falls under inert gas since its outermost electron orbital is full with two electrons.
The answer is 69.253.
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Answers
The right response is that, Using the 0.200 M KBr solution, about 43.31 mL should be mixed with water to create a 0.0495 M solution with a volume of 175.0 mL.
175.0 mL is the final volume needed for a 0.0495 M solution. To achieve this, we can use the dilution formula and calculate the amount of 0.200 M KBr solution necessary.
V2C2 is equivalent to V1C1, as depicted by the formula C1V1 = C2V2.
Where:
Starting at 0.200 M, C1 indicates the concentration of the solution.
Unknown is the designation for the volume of the first solution that will be used.
The diluted solution's final volume is V2, and its ultimate concentration is marked as C2 (0.0495 M).
To determine V1, rewrite the equation as V1 = (C2 * V2) / C1.
Based on these values:
V2 is 175.0 mL, C1 is 0.200 M, and C2 is 0.0495 M.
Enter the corresponding values to establish V1:
= (0.0495 * 175.0) / 0.200 V1 = 43.3125 mL
You would need around 43.31 mL of the 0.200 M KBr solution to make a 0.0495 M solution (rounded to two decimal places) when diluted to 175.0 mL with water.
Learn more about concentration here:
Final answer:
To prepare a 0.0495 M solution of KBr by dilution to 175.0 mL with water, you would need 2.744 mL of 0.200 M KBr solution.
Explanation:
When diluting a solution, the equation M1V1 = M2V2 holds true, where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume. In this case, the final concentration (M2) is 0.0495 M, and the final volume (V2) is 175.0 mL.
Rearranging the equation to solve for V1, the initial volume of the concentrated solution needed, we have:
V1 = (M2 * V2) / M1
Substituting the given values:
V1 = (0.0495 M * 175.0 mL) / 0.200 M
V1 ≈ 43.3125 mL
This means you need 43.3125 mL of the 0.200 M KBr solution to achieve a concentration of 0.0495 M. However, since you have the concentrated solution at 0.200 M, you can dilute it further. The volume you take from the concentrated solution (V1) and the volume of water you add (V_water) should sum up to the final volume of 175.0 mL:
V1 + V_water = 175.0 mL
Rearranging to find V_water:
V_water = 175.0 mL - V1
V_water ≈ 175.0 mL - 43.3125 mL
V_water ≈ 131.6875 mL
So, you would take 43.3125 mL of the 0.200 M KBr solution and dilute it with approximately 131.6875 mL of water to get a total volume of 175.0 mL, resulting in a final concentration of 0.0495 M.
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