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8090 [49]
3 years ago
13

If a substance has a temperature of 273 kelvins, what is its temperature in degrees celsius?

Chemistry
1 answer:
kipiarov [429]3 years ago
5 0
It is at 0 degrees Celsius
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An (blank) determines what will be tested in a scientific experiment
elixir [45]
The answer would be hypotheis since its an theory that isnt proven yet which would involve a scientist to expertiment to make the hypothesis true or valid
6 0
3 years ago
PLEASEEE HELP NOW!!! 60 BRAINLIEST!!
algol13

Answer:

When the net force is balanced

Explanation:

Remeber taking physics and doing this

8 0
3 years ago
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Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine
yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

7 0
3 years ago
Solve fasttt<br><br> C7H17+O2=CO2+H2O<br><br> Balancing Equations
sertanlavr [38]

Answer:

The answer to your question is

                                        4C₇H₁₇  + 45 O₂    ⇒   28 CO₂   +  34H₂O  

Explanation:

Write the equation

                                        C₇H₁₇  +    O₂    ⇒    CO₂   +    H₂O  

Process

1.- Check if the equation is balanced

                                 Reactants            Element              Products

                                        7                          C                           1

                                       17                          H                          2

                                        2                          O                          3

As the number of reactants and products is different, we conclude that the reaction is unbalanced.

2.- Write a coefficient "7" to CO₂   and a coefficient of 17/2 to H₂O

                                      C₇H₁₇  +    O₂    ⇒   7CO₂   +  \frac{17}{2}H₂O  

                                 Reactants            Element              Products

                                        7                          C                           7

                                       17                          H                          17

                                        2                          O                          51/2

3.- Write a coefficient of 45/2 to the O₂, and multiply all the equation by 2.

                         4C₇H₁₇  + 45 O₂    ⇒   28 CO₂   +  34H₂O  

                  Reactants            Element              Products

                        28                          C                        28

                        68                          H                        68

                        90                          O                        90

5 0
3 years ago
Ignore the answer I have selected but please help!! Ty in advance, also please explain
weeeeeb [17]

Answer:

Iodine is most reactive because it is very close to having a "full shell" which is 8 electrons so they are "eager" to gain the last electron to became balanced, so that makes it the most reactive. Hope that helps:)

Explanation:

4 0
3 years ago
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