Silver's chemical symbol is Ag, and it has a 1+ charge. Nitrogen's chemical symbol is N, and it has a 3- charge. For silver and nitrogen to form a compound, their charges must cancel out to zero. Therefore, we would need 3 silver atoms for each nitrogen atom to balance the charges out. The formula for the silver nitride compound would be Ag3N.
Answer:

Explanation:
An empirical formula shows the smallest whole-number ratio of the atoms in a compound.
So, we must calculate this ratio. Since we are given the amounts of the elements in moles, we can do this in just 2 steps.
<h3>1. Divide </h3>
The first step is division. We divide the amount of moles for both elements by the <u>smallest amount of moles</u>.
There are 0.300 moles of sulfur and 0.900 moles of oxygen. 0.300 is smaller, so we divide both amounts by 0.300
- Sulfur: 0.300/0.300= 1
- Oxygen: 0.900/0.300= 3
<h3>2. Write Empirical Formula</h3>
The next step is writing the formula. We use the numbers we just found as the subscripts. These numbers go after the element's symbol in the formula. Remember sulfur is S and there is 1 mole and oxygen is O and there are 3 moles.
This formula is technically correct, but we typically remove subscripts of 1 because no subscript implies 1 representative unit.

Answer:
0.049 mol/L.s
Explanation:
The decomposition of hydrogen peroxide is:

![Rate = -\dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{ 2 \Delta [H_2O_2]}{\Delta t}](https://tex.z-dn.net/?f=Rate%20%3D%20-%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%202%20%20%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D)
The rate of decomposition reaction = the rate of formation of
= 0.098 mol/L.s
∴
Rate of formation of


= 0.049 mol/L.s
Answer:
B
Explanation:
First of all it is important to know that a half filled orbital is particularly stable. In phosphorus all the electrons occur singly in the 3p sublevel minimizing inter electronic repulsion hence it is more difficult to remove an electron from this energetically stable arrangement. In sulphur, electrons are paired in one of the 3p orbitals thereby lowering the energy of that level due to instability caused by interelectronic repulsion between two electrons in the same orbital.
Answer:
B - What we change
Explanation:
Dependent Variable - What we measure
Control Variable - what stays the same
Conclusion - what we conclude
<em>Hope</em><em> </em><em>this</em><em> </em><em>can</em><em> </em><em>Help</em><em>!</em>
<em>:</em><em>D</em>