<h3>
In reference to the diagram attached </h3><h3>
Answer:</h3>
one battery, two switches, and three light bulbs
Explanation:
- A circuit is an arrangement that shows the flow of electrons from a current source such as a battery.
- The diagram shows a circuit that contains one battery, two switches, and three light bulbs.
- The switches serves to close the circuit when necessary.
- The battery supplies current to the circuit.
- The light bulbs are used for lighting.
Answer:
A
Explanation:
rest are nonmetals and they are not shiny
We are given with the initial volume of the substance and the molarity. The first thing that needs to be done is to multiply the equation in order to obtain the number of moles such as shown below.
number of moles = (40 mL) x (1 L / 1000 mL) x (0.3433 moles / L)
number of moles = 0.013732 moles
To get the value of the molarity of the diluted solution, we divide the number of moles by the total volume.
molarity = (0.013732 moles) / (750 mL / 1000 mL/L) = 0.0183 M
Similarly, we can solve for the molarity by using the equation,
M₁V₁ = M₂V₂
Substituting the known values in the equation,
(0.3433 M)(40 mL) = M₂(750 mL)
M₂ = 0.0183 M
From the given balanced equation we have find out the amount (in gm) of Ag formed from 5.50 gm of Ag₂O.
2Ag₂O(s) → 4Ag (s) + O₂ (g)
We know, molecular mass of Ag₂O= 231.7 g/mol, and atomic mass of Ag= 107.8 g/mol. Given, mass of Ag₂O=5.50 gm. Number moles of Ag₂O=
= 0.0237 moles.
From the balanced chemical reaction we get 2 (two) moles of Ag₂O produces 4 (four) moles of Ag. So, 0.0237 moles of Ag₂O produces 
moles=0.0474 moles of Ag= 0.0474 X 107.8 g of Ag=5.11g Ag.
Therefore, 5.50 g Ag₂O produces 5.11 g of Ag as per the given balanced chemical reaction.
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
