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katrin2010 [14]
3 years ago
15

Can science be disproven by philosophy or religion

Chemistry
1 answer:
bagirrra123 [75]3 years ago
7 0
Well, without any bias or my own opinion yes and no, I personally think they cancel each other out in the eyes of scientists some religious events seem out there, but in the eyes of religious leaders they feel that the theory of evolution and such things are a lie because of Adam and Eve. So it really depends on which side of the argument your on and what that entails

P.S. I don’t mean to upset you or anybody I’m just giving my view on the outside from the situation and this is NOT my opinion just simply stating what I see :)
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The equilibrium constant for the formation of ammonia from nitrogen and hydrogen is 1.6 × 102. what is the form of the equilibri
Nimfa-mama [501]

Answer: The expression for equilibrium constant is \frac{[NH_3]^2}{[H_2]^3[N_2]}

Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as k_c

For a general reaction:

aA+bB\rightleftharpoons cC+dD

The equilibrium constant is written as:

k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}

Chemical reaction for the formation of ammonia is:

N_2+3H_3\rightleftharpoons 2NH_3

k_c=1.6\times 10^2

Expression for k_c is:

k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}

1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}

8 0
3 years ago
What is the difference between S-32 and S-36?
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Answer:

its easy ask to chrome or search in yt ;)

8 0
2 years ago
I need this please my brain hurts
GarryVolchara [31]
8.4 grams. I think but I’m not 100% sure
5 0
3 years ago
How many atoms of Carbon are found on the PRODUCTS side?
Semenov [28]
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5 0
3 years ago
The freezing point of ethanol, CH3CH2OH, is -117.300 °C at 1 atmosphere. Kf(ethanol) = 1.99 °C/m
MAXImum [283]

Answer : The molecular weight of this compound is 891.10 g/mol

Explanation :  Given,

Mass of compound = 12.70 g

Mass of ethanol = 216.5 g

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=i\times T_f\times\frac{\text{Mass of compound}\times 1000}{\text{Molar mass of compound}\times \text{Mass of ethanol}}

where,

\Delta T_f = change in freezing point

T_f^o = temperature of pure ethanol = -117.300^oC

T_f = temperature of solution = -117.431^oC

K_f = freezing point constant of ethanol = 1.99^oC/m

i = van't hoff factor = 1   (for non-electrolyte)

m = molality

Now put all the given values in this formula, we get

(-117.300)-(-117.431)=1\times 1.99^oC/m\times \frac{12.70g\times 1000}{\text{Molar mass of compound}\times 216.5g}

\text{Molar mass of compound}=891.10g/mol

Therefore, the molecular weight of this compound is 891.10 g/mol

7 0
3 years ago
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