What is the atomic number, icon, mass of potassium?

This should be EASY!

Answer:

²³⁵₉₂U

Explanation:

The right symbol for the nuclei is ²³⁵₉₂U;

 For this chemical element, the atom is represented by the symbol U;

 To the superscript before the symbol is the mass number

 The subscript before the symbol is the atomic number

Mass number is the number of protons and neutrons

Atomic number is the number of protons.

Answer:

0.88g

Explanation:

The reaction equation:

         2NaI  + Cl₂   →    2NaCl  + I₂

Given parameters:

Mass of Sodium iodide = 2.29g

Unknown:

Mass of NaCl = ?

Solution:

To solve this problem, we work from the known to the unknown.

First find the number of NaI from the mass given;

   Number of moles  =  

          Molar mass of NaI  = 23 + 126.9  = 149.9g/mol

 Now insert the parameters and solve;

     Number of moles  =   = 0.015mol

So;

 From the balanced reaction equation;

            2 moles of NaI produced 2 moles of NaCl

          0.015mole of NaI will produce 0.015mole of NaCl

Therefore;

      Mass  = number of moles x molar mass

      Molar mass of NaCl = 23 + 35.5  = 58.5g/mol

Now;

      Mass of NaCl  = 0.015 x 58.5  = 0.88g

Answer: as the larger quantity

Explanation:

Binary Solution is a homogeneous mixture of two components called as solute and solvent.

Solute is the component which is present in smaller proportion and is solid for solid in liquid solution.

Solvent is the component which is present in larger proportion and is liquid for solid in liquid solution.

Thus solvent is  is usually referred to as the component of a solution which is present as the larger quantity.

Answer:

-1215.9J is the work done

Explanation:

It is possible to find work done in the change of volume of a gas at constant pressure using:

W = -P×ΔV

Where W is work, P is pressure and V is change in volume}

Replacing:

W = -6atm×(5L-3L)

W = -12atmL

As 1atmL = 101,325J, work done in joules is:

-12atmL ×(101.325J / atmL) = -1215.9J is the work done

The value of work done when a volume increases from 3 liters to 5 liters at 6 atm of pressure is -1215.9 Joules.

How do we calculate workdone?

Workdone on any boby or by the body will be calculated as:

W = -P×ΔV, where

W = workdone

P = applied or exerted pressure = 6 atm

ΔV = change in volume due to workdone = (5-3) L

Negative sign in the formula shows that work is done on the opposite side of the pressure or volume.

On putting all these values on the above equation, we get

W = -6atm × (5L-3L)

W = -12 atmL

We know that, 1 atmL = 101,325 J,

So, workdone in joules will be written as:

-12 atmL × (101.325J / atmL) = -1215.9 J

Hence, -1215.9 J is the workdone.

To know more about workdone, visit the below link:
brainly.com/question/18363506

Answer:

7.5 is the answer. You have to move the decimal 2 places to the right

Explanation:

Answer:

1.33%

Explanation:

In an aqueous solution, a weak acid such as acetic acid, will be in equilibrium with its conjugate base, acetate ion, thus:

CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂⁻(aq )

Where dissociation constant, ka, is defined as the ratio of concentrations of products and reactants:

Ka = 1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

H₂O is not taken into account in the equilibrium because is a pure liquid

When a solution of acetic acid becomes to equilibrium, the original concentration of the acid decreases producing more H₃O⁺ and CH₃CO₂⁻.

The concentrations at equilibrium when a 0.100M solution of acetic acid reaches this state, is:

[CH₃CO₂H] = 0.100M - X

[H₃O⁺] = X

[CH₃CO₂⁻] = X

Where X is reaction coordinate.

Replacing in Ka expression:

1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

1.8x10⁻⁵ = [X] [X] / [0.100M - X]

1.8x10⁻⁶ - 1.8x10⁻⁵X = X²

1.8x10⁻⁶ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.00135 → False solution. There is no negative concentrations.

X = 0.00133 → Right solution.

That means concentration of acetate ion is:

[CH₃CO₂⁻] = 0.00133M.

Now, percent ionization is defined as 100 times the ratio between weak acid that is ionizated, [CH₃CO₂⁻] = 0.00133M, per initial concentration of the acid, [CH₃CO₂H] = 0.100M. Replacing:

% Ionization = 0.00133M / 0.100M × 100 =

1.33%