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DochEvi [55]
3 years ago
9

Are the particles in a solid in constant motion, or are they "frozen in place?"

Chemistry
2 answers:
babymother [125]3 years ago
4 0
Particles that are in a solid are tightly locked in place so the answer to your question would be  "Frozen in place"
Tems11 [23]3 years ago
3 0
The molecules are frozen in place but still vibrate
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A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions
Mandarinka [93]

Answer:

\boxed{\text{-55.8 kJ/mol NaOH}}

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\begin{array}{cccl}\text{Heat from neutralization} & + &\text{Heat absorbed by water} & = 0\\q_{1} & + & q_{2} & =0\\n\DeltaH & + & mC\Delta T & =0\\\end{array}

Data:

V(base) = 100.0 mL; c(base) = 0.300 mol·L⁻¹

V(acid) = 100.0 mL; c (acid) = 0.300 mol·L⁻¹  

        T₁ = 35.00 °C;          T₂ = 37.00 °C

Calculations:

(a) q₁

n_{\text{NaOH}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}\\\\n_{\text{HNO}_{3}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}

We have equimolar amounts of NaOH and HNO₃

n = 0.0300 mol

q₁ = 0.0300ΔH

(b) q₂

 V = 100.0 mL + 100.0 mL = 200.0 mL

m = 200.0 g

ΔT = T₂ - T₁ = 37.00 °C – 35.00 °C = 2.00 °C

q₂ = 200.0 × 4.184 × 2.00 = 1674 J

(c) ΔH

0.0300ΔH + 1674 = 0

          0.0300ΔH = -1674

                      ΔH = -1674/0.0300

                      ΔH = -55 800 J/mol

                      ΔH = -55.8 kJ/mol

\Delta_{r}H^{\circ} = \boxed{\textbf{-55.8 kJ/mol NaOH}}

6 0
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