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2 years ago

At the atomic level, are synthetic medicines different from natural medicines? Explain your answer.

Chemistry

2 answers:

nordsb [41]2 years ago

7 0

Answer:

Although herbal medicines are less potent in comparison to synthetic drugs in some cases but still these are consider less toxic or having less side effect in contrast to synthetic drugs. The ultimate norm for any medicine (human made or natural) is their non-toxicity, effectiveness, specificity, stability and potency.

Explanation:

m_a_m_a [10]2 years ago

4 0

Answer:

(It's on quizlet btw)

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2 years ago

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What happens when piece of sodium is exposed to moist air?

I am Lyosha [343]

Answer:

Sodium reacts with the oxygen in air to form sodium oxide, and traces of yellowish sodium peroxide. ... Probably, the reaction with atmospheric oxygen will be faster, because I have never seen sodium metal turning whitish on exposure, though it does lose its lustre and and reactivity (just the surface).

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2 years ago

If one body is positively charged and another one is negatively charged free electrons tend to ?

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The positively charged body has less electrons than protons.

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3 years ago

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Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f

Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0

3 years ago

A quantity of liquid methanol, CH3OH, is introduced intoa rigid 3.00-L vessel, the vessel is sealed, and the temperature israise

Tom [10]

Answer:

74,3 grams is the mass of methanol that was initially introduced into the vessel

Explanation:

CH3OH(g) ----> 1 CO(g) + 2 H2(g)

initial: Idk what i have (X) - -

react: initial - react react react

equilibrium: A concentration A conc. 0.426 M

in equilibrum in eq

See that in equilibrium are formed 2 moles of H2 with this concentration, 0,426M and you form 1 mol of CO, so the moles that are formed of H2 are the double of CO, because stoychiometry.

CH3OH(g) ----> 1 CO(g) + 2 H2(g)

initial: Idk what i have (X) - -

react: initial - react react react

equilibrium: A conc. in eq 0,213 M 0.426 M

So we have concentrations of products in equilibrium, and we have Kc, now we can find concentration of reactants in equilibrium

Kc = ([CO] . [H2]*2) / [CH3OH]

6,9X10*-2 = (0,213 . 0,426*2) / [CH3OH]

[CH3OH] = (0,213 . 0,426*2) / 6,9X10*-2

[CH3OH] = 0,560 M

You know that 1 mol which reacted has this concentration in equilibrium, 0,213 M so I can know what's my initial concentration of reactive

Initial - react = equilibrium

initial - 0,213M = 0,560M ---> Initial = 0,773 M

0,773M is initial concentration of CH3OH, but this is molarity, (moles in 1L). My volume is 3 L so

1 L _____ 0,773 moles

3 L _____ 3L . 0,773moles = 2,319 (the moles that i used)

Molar mass CH3OH = 32.06 g/m

Moles . molar mass = grams ---> 2,319 moles . 32,06 g/m = 74,3 g

8 0

2 years ago