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3 years ago

How are repetition and replication alike and different?

1 answer:

6 0

<span>Both provide approaches to confirming the result of experimentation. Repetition can be developed by one scientist or team continually achieving the expected result but replication requires an independent person or team shows thay can arrive at the same answer independently</span>

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the reaction between nitrogen and hydrogen to produce ammonia is described by the following equilibrium reaction. 3h2 n2↽−−⇀2nh3

eimsori [14]

N₂, H₂, NH₃ are present in the reaction mixture when equilibrium has been obtained.

Balanced chemical reaction: 3H₂ + N₂ ⇄ 2NH₃

Hydrogen (H₂ ) and nitrogen (N₂) are reactants.

Ammonia (NH₃) is product of the reaction.

Reaction goes in both direction, both reactants and product are present.

The amount of substance of reactants and products of reaction do not change when chemical reaction is in chemical equilibrium.

In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which not change with time.

Speed of direct and irreversible chemical reaction are equal.

More about equilibrium: brainly.com/question/25651917

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4 0

1 year ago

What everyday things turn green because acid

nekit [7.7K]

Coins or anything copper

Explanation
They are everyday objects that turn green.

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3 years ago

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k

Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

$ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

We have given in the question

$P_1=102\ mmHg$

$T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol$

And $R$ is the Universal Gas Constant.

$R=0.008 314 kJ/Kmol$

$ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165$

Taking inverse log both side we get,

$\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg$

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3 years ago

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ale4655 [162]

Answer:

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2 years ago

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If the only thing we are interested in studying is how the number of particles affects pressure, what variables do we have to be

andre [41]

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3 years ago