13. Nt
14. St
15. St
16. At
17. Nt
Answer: ₉₈²⁵³Cf
253 is a superscript to the left of the symbol, Cf, which represents the mass number, and 98 is a subscript to the left of the same symbol, which represents the atomic number.
Explanation:
1) The alpha decay equation shows that the isotope Fm - 257, whose nucleus has 100 protons and 157 neutrons, emitted an alpha particle (a nucleus with 2 protons and 2 neutrons).
2) Therefore:
i) the mass number decreased in 4, from 257 to 257 - 4 = 253.
2) the atomic number decreased in 2, from 100 to 100 - 2 = 98.
3) Hence the formed atom has atomic number 98, which is californium, Cf, and the isotope is californium - 253.
4) The item that completes the given alpha decay reaction is:
₉₈²⁵³ Cf.
5) The complete alfpha decay reaction is:
₁₀₀²⁵⁷ Fm → ₉₈²⁵³Cf + ₂⁴He
You can verify the mass balance:
257 = 253 + 4, and
100 = 98 + 2
The formula that would represent an ionic compound that is composed of calcium and iodide ions is CaI2
Explanation
Ionic compound CaI₂ is formed when Calcium form cation ( <em>a positively charged ion</em>) by losing 2 electrons while two iodine atoms form anion ( <em>a negatively charged ion</em>) by gaining one electron each.
When writing down formula of ionic compound, the formula of cation is written first followed by anion formula. therefore Ca is written first followed by I.
The numeric subscript 2 after I(iodine) indicate that 2 atoms of iodine are involved in bonding.
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Here is the complete question.
Benzalkonium Chloride Solution ------------> 250ml
Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.
Sig: Dilute 10ml to a liter and apply to affected area twice daily
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?
(A) 1700 mL
(B) 29.4 mL
(C) 17 mL
(D) 294 mL
Answer:
(B) 29.4 mL
Explanation:
1 L = 1000 mL
1:200 solution implies the 
in 200 mL solution.
200 mL of solution = 1g of Benzalkonium chloride
1000 mL will be 
200mL × 1g = 1000 mL × x(g)
x(g) = 
x(g) = 0.2 g
That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.
∴ 
y(g) = 
y(g) = 5g of benzalkonium chloride.
Now, at 17% concentrate contains 17g/100ml:
∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;
= 
z(mL) = 
z(mL) = 29.41176 mL
≅ 29.4 mL
Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride
