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The answer is Electronegativity
The ml of 12.0 M HCL needed to prepare 905.0 ml of 1.00m hcl is calculated using M1V1 =M2V2 formula
M1= 12.0 M
V1=?
M2= 905.0 ml
V2 = 1.00M
V1 is therefore = M2V2/M1
=905 x 1.00/12.0 = 75.42 ml
Answer:
27.4 g/mol
Explanation:
Assuming the compound is a gas and that it behaves ideally, we can solve this problem by using the <em>PV=nRT formula</em>, where:
- V = 245 mL ⇒ 245 mL / 1000 = 0.245 L
- R = 0.082 atm·L·mol⁻¹·K⁻¹
<u>Inputting the data</u>:
- 1.22 atm * 0.245 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K
<u>Solving for n</u>:
With the <em>calculated number of moles and given mass</em>, we <u>calculate the molar mass</u>:
- 0.334 g / 0.0122 mol = 27.4 g/mol
We need to dilute 0.400 mol of copper (II) sulfate, how do we know, how many weigh of
we have to dilute??
It's simple.
Using a periodic table we can find the molar mass of
Then
now we can replace it
Then we have to dilute 63.84 grams of copper (II) sulfate in 1 L of water to obtain a solution with 0.400M