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Sati [7]
3 years ago
14

Which group of factors are most important in determining the composition of ocean water? a.)temperature, salinity, and density b

.)mass, salinity, and density c,)volume, temperature, and density d.)volume, mass, and density
Chemistry
2 answers:
Elis [28]3 years ago
3 0

Temperature, salinity, and density are the group of factors are most important in determining the composition of ocean water.

a.)temperature, salinity, and density

<u>Explanation:</u>

The three fundamental factors that help in determining the composition of ocean water are temperature, salinity, and density. Temperature, saltiness, salinity, and density influence the thickness of seawater.

Enormous water masses of various densities are significant in the layering of the sea water (increasingly thick water sinks). As temperature builds water turns out to be less thick. As saltiness builds water gets denser. The temperature helps in deciding the pace of vanishing of the ocean.

storchak [24]3 years ago
3 0

Answer:

Temperature, Salinity, and Density

Explanation:

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Explanation:

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What were the first 16 elements known in 1760
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4 0
3 years ago
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Calculate the density (in g/l) of ch4(g) at 75 c and 2.1 atm. (r = 0.08206 latm/molk
vaieri [72.5K]
Answer is: density of methane is 1.176 g/L.<span>
V(CH</span>₄<span>) = 1 L.
T = 75°C = 348.15 K.
p = 2.1 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
n<span> = p·V / R·T.
n</span></span>(CH₄) = 2.1 atm · 1 L / 0.08206 L·atm/mol·K · 348.15 K.
n(CH₄) = 0.0735 mol.
m(CH₄) = n(CH₄) · M(CH₄).
m(CH₄) = 0.0735 mol · 16 g/mol.
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4 0
3 years ago
½O2(g) + H2(g) ⇌ H2O(g)
ira [324]

Answer:

-241.826 kJ·mol⁻¹;  -146.9 J·K⁻¹mol⁻¹; 664.6 J·K⁻¹mol⁻¹; spontaneous

Explanation:

                        ½O₂(g)   +  H₂(g) ⟶ H₂O(g)

ΔHf°/kJ·mol⁻¹:      0                0        -241.826

S°/J·K⁻¹mol⁻¹:   205.0         130.6       188.7

1. ΔᵣH

ΔᵣH = products -reactants = -241.826 -(0 + 0) = -241.826 kJ·mol⁻¹

2. ΔᵣS

ΔᵣS = products - reactants = 188.7 - (205.0 + 130.6) = 188.7 - 335.6 = -146.9 J·K⁻¹mol⁻¹

3. ΔS(univ)

\begin{array}{rcl}\Delta S_{\text{univ}} &=& \Delta S_{\text{sys}}  +\Delta S_{\text{surr}}\\\\ &=& \Delta S_{\text{sys}}  -\dfrac{\Delta H_{\text{sys}}}{T}\\\\& = & -146.9 - \dfrac{-241826}{298}\\\\& = & -146.9 + 811.5\\& = & \mathbf{664.6 \,\, J\cdot K^{-1}mol^{-1}}\\\end{array}

4. Spontaneity

\begin{array}{rcl}\Delta G &=& \Delta H - T\Delta S\\& = & -241.826 - 298 \times (-0.1469)\\& = & -241.826 + 43.776\\& = &  \textbf{-198.050 kJ}\cdot\textbf{mol}^{\mathbf{-1}}\\\end{array}

ΔG is negative, so the reaction is spontaneous.

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