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seropon [69]
3 years ago
13

What the answer I’m on a midterm and can’t seem to find it help

Chemistry
1 answer:
Sonja [21]3 years ago
8 0

Answer:

Sorry I’m not rlly sure but maybe the 2nd or the last

Explanation:

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Natural vs. Synthetic
myrzilka [38]

Answer:

Natural chemicals are produced by nature without any human intervention. Synthetic chemicals are made by humans using methods different than those nature uses, and these chemical structures may or may not be found in nature. Stay safe .

8 0
3 years ago
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OK! this is for the weebs :D
maw [93]

Answer:

Explanation:3

4 0
3 years ago
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What is the pH if the [H+] concentration is 3 x10^-13​
Shkiper50 [21]

Answer:

pH = 12.52

Explanation:

Given that,

The [H+] concentration is 3\times 10^{-13}.

We need to find its pH.

We know that, the definition of pH is as follows :

pH=-log[H^+]

Put all the values,

pH=-log[3\times 10^{-13}]\\\\pH=12.52

So, the pH is 12.52.

6 0
3 years ago
Calculate the enthalpy of the reaction below (∆Hrxn, in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g).
nalin [4]

The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

The bond energies data is given as follows:

BE  for C≡O  = 1072 kJ/mol

BE for Cl-Cl = 242 kJ/mol

BE for C-Cl = 328 kJ/mol

BE for C=O = 766 kJ/mol

The enthalpy change for the reaction is given as :

ΔHr×n = ∑H reactant bond - ∑H product bond

ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )

ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )

ΔHr×n = 1314 - 1422

ΔHr×n = - 108 kJ

Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

To learn more about enthalpy here

brainly.com/question/13981382

#SPJ1

7 0
1 year ago
What is the pressure in mm hg of a 0.025 mole sample of co2 at 350 k in a 200 l container?
Leni [432]

Answer:

0.65mmHg

Explanation:

Take R =8.314J/mol/K

V=200dm3

PV=nRT

P=0.025×8.314×623/200

=0.65mmHg

7 0
1 year ago
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