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Bess [88]
3 years ago
15

Why is it important for scientist to be repeatable?

Chemistry
2 answers:
shutvik [7]3 years ago
8 0

Answer:

It is important for scientists to be repeatable so it lets them see patterns and trends in their results. This is affirmative for their work, making it stronger and better able to support their claims. This helps maintain The integrity of data.

Repeating an experiment more than once helps determine if the data was a fluke, or represents the normal case. It helps guard against jumping to conclusions without enough evidence. The number of repeats depends on many factors, including the spread of the data and the availability of resources

Explanation:

Margarita [4]3 years ago
3 0
It is important for scientists to be repeatable so it lets them see patterns and trends in their results. This is affirmative for their work, making it stronger and better able to support their claims. This helps maintain integrity of data.
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Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr
zysi [14]

<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

N_2+2O_2\rightarrow 2NO_2

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]

We are given:

\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = \frac{121.77}{1}\times 1.90=231.36 J/K

Hence, the value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0
3 years ago
In an energy diagram for a chemical reaction, the reactants are always plotted to the left of the products.
Dominik [7]

the answer is false at lest that is what i know

8 0
3 years ago
Read 2 more answers
Can a mini fridge still be used if you break the freezer
Romashka [77]
The fridge part can, just not the freezer, I think.
8 0
3 years ago
calculate the speed of an electron if its de broglie wavelength is twice its displacement in one second
valkas [14]

Answer:

The speed of an electron is 0.01908 m/s.

Explanation:

De-Broglie wavelength is given by:

\lambda=\frac{h}{mv}

where,  \lambda = wavelength of a particle

h = Planck's constant = 6.626\times 10^{-34}Js

m = mass of particle

v = velocity of the particle

Velocity of an electron = v

Mass of an electron = 9.1\times 10^{-31}kg

Wavelength of electron is twice the displacement in seconds which is velocity of an electron.

Then.wavelength of an electron = \lambda =2v

\lambda =2v=\frac{6.626\times 10^{-34}Js}{9.1\times 10^{-31}kg\times v}

v = 0.01908 m/s

The speed of an electron is 0.01908 m/s.

5 0
3 years ago
Can sombody pls help me with this​
luda_lava [24]
Wouldn’t it be half of each? For 36 I guess is 18 and 54 will be 27, (NOT SURE)
6 0
2 years ago
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