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densk [106]
3 years ago
12

7m-11=10 como se acomoda pls ayuda​

Chemistry
2 answers:
Verdich [7]3 years ago
7 0
La respuesta es m-3
(3)
Reptile [31]3 years ago
3 0

Answer:

3

Explanation:

7m-11=10

quitar 11 de ambos lados

7m -11+11 = 10+11

7m=21

7/7m=21/7

m=3

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When 1.045 g of CaO is added to 50.0 mL of water at 25.0 °C in a calorimeter, the temperature of the water increases to 32.3 °C.
Rus_ich [418]

Answer:

1.71 kJ/mol

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

\Delta H=m\times C\times \Delta T

Where,  

\Delta H  is the enthalpy change

m is the mass

C is the specific heat capacity

\Delta T  is the temperature change

Thus, given that:-

Mass of CaO = 1.045 g

Specific heat = 4.18 J/g°C

\Delta T=32.3-25.0\ ^0C=7.3\ ^0C

So,  

\Delta H=1.045\times 4.18\times 7.3\ J=31.88713\ J

Also, 1 J = 0.001 kJ

So,  

\Delta H=0.03189\ kJ

Also, Molar mass of CaO = 56.0774 g/mol

Moles=\frac{Mass}{Molar\ mass}=\frac{1.045}{56.0774}\ mol=0.01863\ mol

Thus, Enthalpy change in kJ/mol is:-

\Delta H=\frac{0.03189}{0.01863}\ kJ/mol=1.71\ kJ/mol

8 0
2 years ago
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
what might cause the goldfish population to drop if the temperature moves out of its normal range? Name three things
Slav-nsk [51]

Answer:

Appearing disoriented, such as swimming upside down.

Leaving food uneaten.

White spots on fins or body.

Discolored gills.

Explanation:

Gold Fish are unable to regulate their body temperature, so they are influenced by the temperature around them. If the water is warm, gold fish metabolism accelerates, feeding and respiration increases, and there is a general increase in movement. If the water is cooler, fish become lethargic and tend to be inactive.

8 0
3 years ago
Which muscles does the brain control
kozerog [31]

Answer:

Its Cerebrum a muscle that the brain control

8 0
3 years ago
Help me with my work please
blsea [12.9K]

Answer:

i am pretty sure it is compound

8 0
3 years ago
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