Answer:
C. chains of alternating silicon and oxygen atoms with attached organic groups
Explanation:
It contains Silicone-Oxygen backbone as found in Silicone Dioxide (silica) but it also has organic groups attached to this backbone with Carbon-Silicone attachments. This whole arrangement of Silicone, Oxygen and Attached organic groups make up Silicones.
Its used for producing ferrite magnets and refining zinc. its salt causes vivid color crimson flames to produce flares
For all question, all you need to use is the mole-mole ratio.
a) 25 moles C2H6O (3 moles O2/ 1 mol C2H6O)= 75 moles O2
b) 30 moles O2 (1 moles C2H6O/ 3 moles O2)= 10 moles C2H6O
c) 23 moles CO2 (3 moles O2/ 2 moles CO2) = 34.5 moles O2
d) 41 moles H2O ( 1 moles C2H6O/ 3 moles H2O= 13.7 moles C2H6O
Answer:
Hence the correct option is an option (b) Sr4, Cl,Br−,Na+.
Explanation:
Bromine and chlorine belong to an equivalent group. As we go down the group the dimensions increases which too there's a charge on the bromine atom. therefore the size of the Br- is going to be larger in comparison to the chlorine atom.
Sr atom is within the second group, and also it's below the above-mentioned atoms.so Sr is going to be the larger one among all the atoms.
Sodium and chlorine belong to an equivalent period .size decrease from left to right. but due to the charge on sodium its size decreases and there's an opportunity that Na+ size could be adequate for Cl.
Here we finally assume that two atoms are of an equivalent size (Na+ and Cl) which are less in size compared to the opposite two(Sr and Br-) during which one is greater (Sr)and the opposite is smaller(Br-).
Answer:
The amount of drug left in his body at 7:00 pm is 315.7 mg.
Explanation:
First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:

Where:
λ: is the decay constant = 
: is the half-life of the drug = 3.5 h
N(t): is the quantity of the drug at time t
N₀: is the initial quantity
After 90 min and before he takes the other 200 mg pill, we have:

Now, at 7:00 pm we have:

Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).
I hope it helps you!