Answer:
S= 2(1) = 2
O= 2(4) = 8
Na= 2(2) = 4
Explanation:
The given compound is:
2Na₂SO₄
An element is a distinct substance that cannot be split up into simpler substances.
So;
Number of atoms of elements here are:
S= 2(1) = 2
O= 2(4) = 8
Na= 2(2) = 4
The given question is incomplete. The complete question is:
A chemist prepares a solution of barium chloride by measuring out 110 g of barium chloride into a 440 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mole per liter of the chemist's barium chloride solution. Round your answer to 3 significant digits.
Answer: Concentration of the chemist's barium chloride solution is 1.20 mol/L
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

where,
n = moles of solute
= volume of solution in L
moles of
(solute) = 
Now put all the given values in the formula of molality, we get

Therefore, the molarity of solution is 1.20 mol/L
Use blue litmus paper. This is an indicator that can safely determine whether it is a base or an acid by changing color in response to the substance. This color indicates whether it is an acid or a base. Refer to the pH scale to see if the substance is basic or acidic.
So platinum is a transition metal. In general transition metals are reducers, which means they can give the electrons they have, to the sodium atoms. Also in chemistry we look at sub orbitals rather that shells(2,8,8). So due to the energy from heat, the d orbital split as electrons move to a higher energy level. Some of the electrons are given to the sodium ions and therefore the flame changes colour to yellow.
The excitation of the electrons is caused by them getting energy and so moving up an energy level. This energy is released and the electron returns to it's original state. The energy released, however, does not release in the same direction, but in different/various directions. Therefore the colour of the light changes as some energy is released in the surrounding.