The early atmosphere was probably mostly carbon dioxide, with little or no oxygen. <span>The proportion of oxygen went up because of </span>photosynthesis. The photosynthesis was conducted from <span>tiny organisms.
</span><span>cyanobacteria, or blue-green algae. </span><span>
They </span>used sunshine, water and carbon dioxide to produce carbohydrates and, yes, oxygen. This change to the atmosphere was very important because the <span>breathable air we enjoy today was created.</span>
Answer:
c = 0.898 J/g.°C
Explanation:
1) Given data:
Mass of water = 23.0 g
Initial temperature = 25.4°C
Final temperature = 42.8° C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Specific heat capacity of water is 4.18 J/g°C
ΔT = 42.8°C - 25.4°C
ΔT = 17.4°C
Q = 23.0 g × × 4.18 J/g°C × 17.4°C
Q = 1672.84 j
2) Given data:
Mass of metal = 120.7 g
Initial temperature = 90.5°C
Final temperature = 25.7 ° C
Heat released = 7020 J
Specific heat capacity of metal = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 25.7°C - 90.5°C
ΔT = -64.8°C
7020 J = 120.7 g × c × -64.8°C
7020 J = -7821.36 g.°C × c
c = 7020 J / -7821.36 g.°C
c = 0.898 J/g.°C
Negative sign shows heat is released.
The pH of the unknown solution is 3.07.
<u>Explanation:</u>
<u>1.Find the cell potential as a function of pH</u>
From the Nernst Equation:
Ecell=E∘cell−RT /zF × lnQ
where
R denotes the Universal Gas Constant
T denotes the temperature
z denotes the moles of electrons transferred per mole of hydrogen
F denotes the Faraday constant
Q denotes the reaction quotient
Substitute the values,
E∘cell=0 lnQ=2.303logQ
E0cell=−2.30/RT /zF × log Q
Solving the equation,
<u>2. Find the Q value</u>
Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant
Q=[H+]^2×1/1×1=[H+]2
Taking the log
logQ= log[H+]^2=2log[H+]=-2pH
From the formula,
Ecell=−2.303RT /zF× logQ
E cell= 2.303 × 8.314 CK mol (inverse) × 298.15
K × 2pH /2×96 485 C⋅mol
( inverse)
E cell= 0.0592 V × pH
<u>3. Finding the pH value</u>
E cell= 0.0592 V × pH
pH = E cell/ 0.0592 V= 0.182V/ 0.0592V
pH=3.07
The pH of the unknown solution is 3.07.
The answer is dissolved in water, because in chemistry, (aq) is shorthand for aqueous solution.
The Sun<span> is a main-sequence star, and thus generates its energy by </span>nuclear fusion<span> of hydrogen nuclei into helium. In its core, the </span>Sun<span> fuses 620 million metric tons of hydrogen each second.</span>
