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3 years ago

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What is medical chemistry?

2 answers:

Dahasolnce [82]3 years ago

5 0

Answer:

Medicinal chemistry is a discipline that encloses the design, development, and synthesis of pharmaceutical drugs. The discipline combines expertise from chemistry, especially synthetic organic chemistry, pharmacology, and other biological sciences.

Explanation:

Its called medicinal chemistry

pogonyaev3 years ago

4 0

Explanation:

Medicinal chemistry is a discipline that encloses the design, development, and synthesis of pharmaceutical drugs. The discipline combines expertise from chemistry, especially synthetic organic chemistry, pharmacology, and other biological sciences.

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Which mass of urea CO(NH2)2 contain the same mass of nitrogen as 101.1g of potassium nitrate?

Sliva [168]

M(KNO₃)=101.1 g/mol
M(CO(NH₂)₂)=60.1 g/mol

m(N)=M(N)m(KNO₃)/M(KNO₃)

m(N)=2M(N)m(CO(NH₂)₂)/M(CO(NH₂)₂)

2m(CO(NH₂)₂)/M(CO(NH₂)₂)=m(KNO₃)/M(KNO₃)

m(CO(NH₂)₂)=M(CO(NH₂)₂)m(KNO₃)/(2M(KNO₃))

m(CO(NH₂)₂)=60.1*101.1/(2*101.1)=30.05 g

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Name the following compound:<br><br> CH3 – CH2 – CH2 – O – CH2 – CH2 – CH2 – CH3

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THE COMPOUND IS: ethnoxybutane

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3 years ago

What is the name of the compound BrF

Rashid [163]

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Read 2 more answers

Convert 145 meters to kilometers.

Sphinxa [80]

Divide by 1000

145÷1000=0.145

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1 year ago

2. Consider the reaction 2NO(g) + O2(g) → 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is r

vredina [299]

Answer :

(a) The rate of $NO_2$ formed is, 0.066 M/s

(b) The rate of $O_2$ formed is, 0.033 M/s

Explanation : Given,

$\frac{d[NO]}{dt}$ = 0.066 M/s

The balanced chemical reaction is,

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate of disappearance of $NO$ = $-\frac{1}{2}\frac{d[NO]}{dt}$

The rate of disappearance of $O_2$ = $-\frac{d[O_2]}{dt}$

The rate of formation of $NO_2$ = $\frac{1}{2}\frac{d[NO_2]}{dt}$

As we know that,

$\frac{d[NO]}{dt}$ = 0.066 M/s

(a) Now we have to determine the rate of $NO_2$ formed.

$\frac{1}{2}\frac{d[NO_2]}{dt}=\frac{1}{2}\frac{d[NO]}{dt}$

$\frac{d[NO_2]}{dt}=\frac{d[NO]}{dt}=0.066M/s$

The rate of $NO_2$ formed is, 0.066 M/s

(b) Now we have to determine the rate of molecular oxygen reacting.

$-\frac{d[O_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}$

$\frac{d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s$

The rate of $O_2$ formed is, 0.033 M/s

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3 years ago