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Fofino [41]
4 years ago
8

If the density of pure water is 0.9922 g/mL at 40 ºC, calculate its theoretical molarity at that temperature. Report to 4 sig fi

gs. (Only include numerical portion of your answer - do not include units)
Chemistry
1 answer:
OleMash [197]4 years ago
3 0
Answer is: theoretical molarity of water is 55.1222 mol/L.<span>
d(H</span>₂O) = 0.9922 g/mL.
M(H₂O) = 2 · Ar(H) + Ar(O) · g/mol.
M(H₂O) = 2 + 16 · g/mol = 18 g/mol.
c(H₂O) = d(H₂O) ÷ M(H₂O).
c(H₂O) = 0.9922 g/mL ÷ 18 g/mol.
c(H₂O) = 0.0551 mol/mL.
c(H₂O) = 0.0551 mol/mL · 1000 mL/L = 55.1222 mol/L.
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1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction
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4 0
3 years ago
If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t
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\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

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Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

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m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]      ......(1)

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c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

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T_{final}=49^oC

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