34% I believe
Hope this helps!
STSN
BY CHECKING THE REACTIVITY OF AN ELEMENT WHICH IS MOST REACTIVE OR NOT AND YOU STUDY TYPE OF CHEMICAL REACTION IN 1 CH AND YOU CHECK THE REACTIVITY OF ELEMENTS IN 3 CH METALS AND NON METALS PAGE NO 45 IN NCERT BOOK
<h3>Answer </h3>
After another 5730 years ( three half lives or 17190 years) 17.5 /2 = 8.75mg decays and 8.75g remains left. after three half lives or 17190 years, 8.75 g of C-14 will be
Explanation:
hope this help
Answer:
0.42 M
Explanation:
The reaction that takes place is:
- Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)
First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:
(200 mL = 0.200L)
- 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄
Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:
- 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂
Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:
- 0.224 mol - 0.14 mol = 0.085 mol
Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:
- 0.085 mol / 0.200 L = 0.42 M