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RSB [31]
3 years ago
8

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!! 100 points!

Chemistry
1 answer:
Elenna [48]3 years ago
8 0

Answer:

Explanation:  Hydrochloric Acid (HCl) Hydrochloric acid is the strong acid which is found inside our body in the gastric juice. It helps in the breakdown of all the potatoes and pizzas we eat and turns down the enzymes, which change the food particles into protein.LSD is made by creating a fungus on seeds of rye or morning glory. A hallucinogenic drug, capable of changing the way people look at, think about, and experience the world around them. The power of this drug makes it popular, as does its portability.

An acid is a chemical species that donates protons or hydrogen ions and/or accepts electrons. Most acids contain a hydrogen atom bonded that can release (dissociate) to yield a cation and an anion in water. The higher the concentration of hydrogen ions produced by an acid, the higher its acidity and the lower the pH of the solution.

The word acid comes from the Latin words acidus or acere, which mean "sour," since one of the characteristics of acids in water is a sour taste (e.g., vinegar or lemon juice).

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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
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Answer:

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

Explanation:

NaOH + HNO₃ ⟶ NaNO₃ + H₂O

There are two energy flows in this reaction.

\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
3 years ago
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