This is so frustrating I don’t even understand this
The question is incomplete, here is the complete question.
A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.
<u>Answer:</u> The concentration of copper fluoride in the solution is ![4.90\times 10^{-3}mol/L](https://tex.z-dn.net/?f=4.90%5Ctimes%2010%5E%7B-3%7Dmol%2FL)
<u>Explanation:</u>
To calculate the molarity of solute, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%20%28in%20mL%29%7D%7D)
We are given:
Given mass of copper (II) fluoride = 0.0498 g
Molar mass of copper (II) fluoride = 101.54 g/mol
Volume of solution = 100.0 mL
Putting values in above equation, we get:
![\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20copper%20%28II%29%20fluoride%29%3D%5Cfrac%7B0.0498%5Ctimes%201000%7D%7B101.54%5Ctimes%20100.0%7D%5C%5C%5C%5C%5Ctext%7BMolarity%20of%20copper%20%28II%29%20fluoride%7D%3D4.90%5Ctimes%2010%5E%7B-3%7Dmol%2FL)
Hence, the concentration of copper fluoride in the solution is ![4.90\times 10^{-3}mol/L](https://tex.z-dn.net/?f=4.90%5Ctimes%2010%5E%7B-3%7Dmol%2FL)
The volume that will be occupied at 735 torr and 57 c is 23.12 L
<u><em>calculation</em></u>
- <u><em> </em></u> At STP temperature=273 k and pressure=760 torr
- <u><em> </em></u>by use of combined gas formula
that is P1V1/T1= P2V2/T2
where; P1 =760 torr
T1= 273 K
V1= 18.5 L
P2= 735 torr
T2= 57+273= 330 K
V2=?
- by making V2 the formula of subject
V2= T2P1V1/P2T1
V2= [(18.5L x 330 k x 760 torr)/(735 torr x 273 k)]= 23.12 L
a. pH=2.07
b. pH=3
c. pH=8
<h3>Further explanation</h3>
pH=-log [H⁺]
a) 0.1 M HF Ka = 7.2 x 10⁻⁴
HF= weak acid
![\tt [H^+]=\sqrt{Ka.M}\\\\(H^+]=\sqrt{7.2.10^{-4}\times 0.1}\\\\(H^+]=8.5\times 10^{-3}\\\\pH=3-log~8.5=2.07](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D%5Csqrt%7B7.2.10%5E%7B-4%7D%5Ctimes%200.1%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D8.5%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5CpH%3D3-log~8.5%3D2.07)
b) 1 x 10⁻³ M HNO₃
HNO₃ = strong acid
![\tt pH=-log[1\times 10^{-3}]=3](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-3%7D%5D%3D3)
c) 1 x 10⁻⁸ M HCl
![\tt pH=-log[1\times 10^{-8}]=8](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-8%7D%5D%3D8)
Answer:
The first option: Strontium Fluorate.
Explanation:
because Fluorine and oxygen combines to make fluorate, Strontium stays the same.
p.s: i need help in geo and there's an exam tomorrow.