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Vsevolod [243]
3 years ago
11

Worn away statue Mechanical? or Chemical?

Chemistry
2 answers:
Daniel [21]3 years ago
5 0
It probably depends on what the statue is made of for example:
Stone,wood,etc... But the answer is probably chemical because it is being worn away and chemicals are going on to it.
kompoz [17]3 years ago
3 0
Yes it is chemical because it's acid rain and other stuff that is warring it out
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Plants and animals add carbon dioxide to the atomsphere when they ____ glucose.
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Name the following alkenes using systematic names.
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4 years ago
Aspirin (C9H8O4) is synthesized by reacting salicylic acid (C7H6O3) with acetic anhydride (C4H6O3) according to the reaction: C7
Lunna [17]

Answer:

7.39\times 10^{3}g of acetic anhydride is needed

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So, 1.00\times 10^{4}g of salicylic acid = \frac{1.00\times 10^{4}}{138.121}moles of salicylic acid

Hence, \frac{1.00\times 10^{4}}{138.121}moles of salicylic acid reacts completely with \frac{1.00\times 10^{4}}{138.121}moles of acetic anhydride.

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7 0
4 years ago
A 18.08-g sample of the ionic compound , where is the anion of a weak acid, was dissolved in enough water to make 116.0 mL of so
Oksi-84 [34.3K]

Answer:

a) 129.14 g/mol

b) 8.87

Explanation:

Given that:

mass of the ionic compound [NaA] = 18.08 g

Volume of water = 116.0 mL = 0.116 L

Let the mole of the acid HCl = 0.140 M

Volume of the acid = 500.0 mL = 0.500 L

pH = 4.63

V_{equivalence}_{acid} = 1.00 L

Equation for the reaction can be represented as:

NaA_{(aq)} + HCl_{(aq)} -----> HA_{(aq)} + NaCl_{(aq)

From above; 1 mole of an ionic compound reacts with 1 mole of an acid to reach equivalence point = 0.140 M × 1.00 L

= 0.140 mol

Thus, 0.140 mol of HCl neutralize 0.140 mol of ionic compound at equilibrium

Thus, the molar mass of the sample = \frac{18.08g}{0.140 mole}

= 129.14 g/mol

b) since pH = pKa

Then pKa of HA = 4.63

Ka = 10^{-4.63]

= 2.3*10^{-5}

[A^-]equ = \frac{0.140M*1.00L}{1.00L+0.116L}

= \frac{0.140 mol}{1.116 L}

= 0.1255 M

K_a of HA = 2.3*10^{-5}

K_b = \frac{1.0*10^{-14}}{2.3*10^{-5}}

= 4.35*10^{-10}

                     A_{(aq)}     +     H_2O_{(l)}         \rightleftharpoons     HA_{(aq)}     +     OH^-_{(aq)}

Initial        0.1255                                            0                    0

Change     - x                                                  +  x                 + x

Equilibrium   0.1255 - x                                   x                    x

K_b = \frac{[HA][OH^-]}{[A^-]}

4.35*10^{-10} = \frac{[x][x]}{[0.1255-x]}

As K_b is very small, (o.1255 - x) = 0.1255

x = \sqrt{0.1255*4.35*10^{-10}}

[OH⁻] = x = 7.4 *10^{-6}

But pOH = - log [OH⁻]

= - log [7.4*10^{-6}]

= 5.13

pH = 14.00 = 5.13

pH = 8.87

6 0
3 years ago
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