What is the periodic table a list of

If chlorine gas is bubbles through an aqueous solution of sodium iodide,the result is elemental iodine and aqueous sodium chlori

Mrac [35]

The balanced reaction that describes the reaction of chlorine gas and sodium iodide to produce elemental iodine and sodium chloride in aqueous solution is expressed Cl2+2NaI= I2 + 2NaCl. This kind of reaction is called single replacement reaction where the anion, in this case, is only replaced

3 0

3 years ago

Which statements describe how heat flows in foil? Check all that apply. Heat flows in all directions. Heat flows from left to ri

docker41 [41]

Answer:

A D F

Explanation:

Its right but its not in order But its A D and F

3 0

2 years ago

Do cells divide twice in asexual reproduction, sexual reproduction, or both?

igor_vitrenko [27]

Cells divide in asexual reproduction!

I hope this helps

please mark me as brainliestt!!

8 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 $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

2 years ago

What volume would 0.735 moles of O2 gas occupy at STP

tatiyna

Answer:

16.5 dm³

Explanation:

Data Given:

no. moles of O₂ = 0.735 moles

volume of O₂ = ?

Solution:

Now

we have to find volume of O₂ gas

Formula used for this purpose

No. of moles = Volume / molar volume

where

molar volume at STP for Oxygen (O₂) = 22.4 dm³/mol

No. of moles O₂ = Volume of O₂ / 22.4 dm³/mol . . . . . .(1)

Put values in equation 1

0.735 = Volume of O₂ / 22.4 dm³/ mol

rearrange above equation

Volume of O₂ = 0.735 x 22.4 dm³/ mol

Volume of O₂ = 16.5 dm³

So,

the volume of O₂ at STP is 16.5 dm³

6 0

3 years ago