ans A. 13 to 3
solution:
Total puppies for sale=6
Total puppies left = 2
hence total puppies sold= 6 - 2 = 4
cost of a puppy= $ 104
so cost of 2 puppies = 104 x 4
=416$
Now for kittens:
Total kittens for sale = 12
Total kittens left = 8
hence total kittens sold=12 - 8 =4
cost of a kitten = $24
so cost of 4 kittens= 24x 4=96$
Now ratio of sales of puppies to kittens = 208$/96$
= 13/3 or 13:3 or 13 to 3
I hope this explanation will help you to understand this problem
Answer:
1.07 g
Explanation:
Half-life of Pu-234 = 4.98 hours
Initially present = 45 g
mass remains after 27 hours = ?
Solution:
Formula
mass remains = 1/ 2ⁿ (original mass) ……… (1)
Where “n” is the number of half lives
To find "n" for 27 hours
n = time passed / half-life . . . . . . . .(2)
put values in equation 2
n = 27 hr / 4.98 hr
n = 5.4
Mass after 27 hr
Put values in equation 1
mass remains = 1/ 2ⁿ (original mass)
mass remains = 1/ 2^5.4 (45 g)
mass remains = 1/ 42.2 (45 g)
mass remains = 0.0237 x 45 g
mass remains = 1.07 g
Answer:0.005M
Explanation:
First deduce the oxidation and reduction half equations and from that obtain the balanced redox reaction equation. From that, the number of moles of reacting species are seen from the stoichiometry of the reaction from which the number of moles of oxalate is obtained and substituted to obtain the molar concentration of oxalate.
Solid , solid, partially melted, liquid so the answer is D
