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vivado [14]
3 years ago
15

Which statement about trade during the period of the Roman Empire is correct? A. Trade supplied Romans with exotic goods and hel

ped spread new ideas. B. Trade was confined to food staples transported over short distances. C. Most of the empire's outlying territories refused to trade with Rome. D. A lack of paved, well-maintained overland routes hampered Roman trade.
Chemistry
2 answers:
Lerok [7]3 years ago
6 0
The answer Is A. Tarde supplied Romas with exotic goods.

I did the test to.
densk [106]3 years ago
6 0

Option A i.e Trade supplied Romans with exotic goods and helped spread new ideas is the correct answer.

The Roman Empire was the center of trade between the East and the West. Besides trade of goods, the trade helped Spread new ideas and culture of East and the West in the Rome. So Trade routes were dissemination of new ideas and culture in the past.

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Answer:

3%

Explanation:

Substract the actual error from the final and multiply by 100

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Put the list in chronological order (1–5).
Leokris [45]

Explanation:

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if you were to engineer an everyday solution for conserving water which activity do you think would be the most impactful ?
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If you were to engineer an everyday solution for conserving water which activity do you think would be the most impactful ?
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1 year ago
A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

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2 years ago
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Answer:

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Explanation:

Given data:

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Solution:

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Mathematical relationship:

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Now we will put the values in formula:

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P₂ = 391437.5 atm. K /298.15 K

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