Remember that any intersection of lines is a C, and that the number of hydrogens attached are the necessary to complet the 4 bonds.
1) CH3 - CH (OH) - CH (CH3) -CH3
2) CH3 - O - CH(CH3)-CH2 - CH3
I have used the parenthesis to indicate that the radical inside is in other branch, bonded by a single line -
Answer:
a) AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Explanation:
a) AgNO3 + KI → Ag+ + NO3- + K+ + I-
Ag+ + NO3- + K+ + I- → AgI + KNO3
AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba^2+ + 2OH- + 2H+ + 2NO3-
Ba^2+ + 2OH- + 2H+ + 2NO3- → Ba(NO3)2 + 2H2O
Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → 6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3-
6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3- → Ni3(PO4)2 + 6NaNO3
2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → 2Al^3+ + 6OH- + 6H+ + 3SO4^2-
2Al^3+ + 3OH- + 3H+ + 3SO4^2- → Al2(SO4)3 + 6H2O
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Answer:
You cannot make observations if you are 57 seconds late into the lab.
Explanation:
The atomic nucleus can split by decay into 2 or more particles as a result of the instability of its atomic nucleus due to the fact that radioactive elements possess an unstable atomic nucleus.
Now, the primary particles which are emitted by radioactive elements in order to make them decay are alpha, beta & gamma particles.
The half life equation is;
N_t = N₀(½)^(t/t_½)
Where:
t = duration of decay
t_½ = half-life
N₀ = number of radioactive atoms initially
N_t = number of radioactive atoms remaining after decay over time t
We are given;
t = 57 secs
N₀ = 100 g
Now, half life of Nitrogen-16 from online sources is 7.2 seconds. t_½ = 7.2
Thus;
N_t = 100(1/2)^(57/7.2)
N_t = 0.4139g
We are told that In order to make observations, you require at least .5g of material.
The value of N_t you got is less than 0.5g, therefore you cannot make observations if you are 57 seconds late.
Answer:
0.05 mol
Explanation:
The balanced equation for the reaction that takes place is:
- 2C₂H₂ (g) + 5O₂ (g) → 4CO₂ (g) + 2H₂O (g)
Now we<u> convert 0.10 moles of carbon dioxide (CO₂) into moles of acetylene (C₂H₂)</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 0.10 mol CO₂ *
= 0.05 mol C₂H₂
