Answer:
[NH₃] → 3.24 M
Explanation:
Our solute: Ammonia
Our solvent: Water
Solution's mass = Mass of solute + Mass of solvent
Solution's mass = 15 g + 250 g = 265g
We use density to determine, the volume.
D = mass /volume → Volume = m / D → 265 g /0.974 g/mL = 272.07 mL.
We convert the mL to L → 272.07 mL . 1L /1000mL = 0.27207 L
To determine molarity we need the moles of solute in 1 L of solution.
Moles of solute are: 15g / 17g/mol = 0.882 moles
[NH₃] = 0.882mol /0.27207 L → 3.24 M
Answer:
<h2>4.0 </h2>
Explanation:
The pH of a solution can be found by using the formula
![pH = - log [ { H}^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5B%20%7B%20H%7D%5E%7B%2B%7D%5D)
From the question we have
We have the final answer as
<h3>4.0</h3>
Hope this helps you
Answer:
1. 1
2. 2
3. 6
4. 8
Explanation:
Valence electrons can be defined as the number of electrons present in the outermost shell of an atom. Valence electrons are used to determine whether an atom or group of elements found in a periodic table can bond with others. Thus, this property is typically used to determine the chemical properties of elements.
1. Li is the symbol for the chemical element known as Lithium: it has 1 valence electron. The electronic configuration of lithium is 1s²2s¹
2. Mg is the symbol for the chemical element known as Magnesium: it has 2 valence electrons. The electronic configuration of magnesium is 1s²2s²2p63s²
3. O is the symbol for the chemical element known as Oxygen: it has 6 valence electrons. The electronic configuration of magnesium is 1s²2s²2p⁴.
4. Ne is the symbol for the chemical element known as Neon: it has 8 valence electrons.
Answer:
Isopropylbenzene
Explanation:
If you draw the structure, you can see that there are two methyl groups and in between there.
Adjacent to CH3, there are four neighbouring hydrogens, therefore, n=4+1 = 5. The same is for methyl on other side. For carbon present in benzene ring, there is 2, since there is one hydrogen on benzene per carbon.
Yes. density is a physical property that can be used to identify matter
