Alguien sabe el nombre del compuesto etilisopropilacetileno?

If HCl is a 0.05 M solution and 50 mL is used to titrate the NaOH, what is the molarity of NaOH if the flask contains 100 mL?

Liono4ka [1.6K]

Answer:

Molarity of NaOH = 0.025 M

Explanation:

Given data:

Molarity of HCl = C₁ = 0.05 M

Volume of HCl = V₁= 50 mL

Molarity of NaOH = C₂=?

Volume of NaOH =V₂= 100 mL

Solution:

Formula:

C₁V₁ = C₂V₂

C₁ = Molarity of HCl

V₁ = Volume of HCl

C₂ = Molarity of NaOH

V₂ = Volume of NaOH

Now we will put the values:

C₁V₁ = C₂V₂

0.05 M × 50 mL = C₂ × 100 mL

2.5 M.mL =C₂ × 100 mL

C₂ = 2.5 M.mL /100 mL

C₂ = 0.025 M

7 0

3 years ago

Which is an example of a set point for the body? A. hair color B. type of blood C. thumb crossing style D. amount of blood gluco

Mama L [17]

The Answer Is B Type Of Blood

7 0

3 years ago

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$