One mole of NaCl weighs 58.44 grams
Therefore 1L of one mole of NaCl contains 58.44 grams
1L of 0.2m NaCl contains 58.44x0.2 = 11.69grams
300mL of 0.2m NaCl contains 11.69x0.3 = 3.51 grams
First, you need to find the number of moles of OH⁻ in a 250mL solution of 0.100M OH⁻. To do this, multiply 0.250L by 0.100M to get 0.025mol OH⁻. Then you use the fact that 1 mole of Sr(OH₂)·8H₂O contains 2 moles of OH⁻ which means that 0.0125mol of Sr(OH)₂·8H₂O contains 0.025mol OH⁻ (0.025/2=0.0125). Then to find the amount of Sr(OH)₂·8H₂O is needed you multiply its molar mass (265.76g/mol) by 0.0125mol to get 3.322g.
Therefore you need 3.322g of Sr(OH)₂·8H₂O.
I hope that helps. Let me know if anything is unclear.
Answer:
The answer to your question is: Fluorine, Chlorine, Bromine, Iodine
Explanation:
Electron configuration ns² np⁵
We observe that this electron configuration finishes in "p". Elements from groups IIIA to VIIIA finish their electron configuration in "p".
Elements in Group IIIA finish in p¹
Group IVA finish in p²
Group VA finish in p³
Group VIA finish in p⁴
Group VIIA finish in p⁵
Group VIIIA finish in p⁶
Then, the list of elements with this electron configuration is:
Fluorine, Chlorine, Bromine, Iodine
The number of protons in the nucleus for that element
Temperature <span>is a measure of the average kinetic energy of particles, not just gas. </span>