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valkas [14]
2 years ago
6

Select the formula for the following acid: Carbonic Acid

Chemistry
1 answer:
Nina [5.8K]2 years ago
4 0

Answer: 106

Explanation:

Follow PEMDAS. 7x12 = 84

84 + 13 = 97

97 + 9 = 106

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Structures of compounds people use every day are shown. From which group of unsaturated hydrocarbons is each derived? 2 carbons
Savatey [412]

The group of unsaturated hydrocarbons which 2 carbons are double bonded together, with H bonded to the left, and C H 2 bonded below left, above right, and below right is derived from <u>Alkenes</u>

<h3>What are organic compounds?</h3>

Organic compounds are compounds which contains carbon and hydrogen

Some few classes or organic compounds or hydrocarbons are as follows:

  • Alkanes
  • Alkenes
  • Alkynes
  • Alkanols
  • Alkanoic acid
  • Ketones
  • Esters

So therefore, the group of unsaturated hydrocarbons which 2 carbons are double bonded together, with H bonded to the left, and C H 2 bonded below left, above right, and below right is derived from <u>Alkenes</u>

Learn more about organic compounds:

brainly.com/question/1594044

#SPJ1

3 0
1 year 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
Which of the following procedures is used when refining petroleum? Increase the temperature of the oil. Decrease the temperature
Novosadov [1.4K]
Increase the tempature
6 0
2 years ago
Read 2 more answers
Lithium reacts with bromine (Br2) in a synthesis reaction to produce lithium bromide. Determine the limiting reactant if 25.0 gr
Lunna [17]

Answer: Bromine is the limiting reactant

Explanation:

First of all let's generate a balanced equation for the reaction

2Li + Br2 —> 2LiBr

Molar Mass of Li = 7g/mol

Molar Mass of Br2 = 2x80 = 160g/mol

From the question given, were told that 25g of Li and 25g Br2 were present at the take-off of the reaction. Converting these Masses to mole, we have:

Number of mole of Li = 25/7 = 3.6moles

Number of mole of Br2 = 25/160 = 0.156mol.

To know which is the limiting reactant, we have to compare the ratio of the number of mole of experimental Li and Br2 to that of theoretical Li and Br2

For the experimental yield:

Li : Br2 = 3.6/ 0.156 = 23 : 1

For the theoretical yield:

Li : Br = 2 : 1

From the above, we see clear that Br2 is the limiting reactant because according to the equation( which gives the theoretical yield), for every 2moles of Li, 1mole of Br2 is used up. But this is not so from the experiment conducted as 23moles required 1mole of Br2.

4 0
3 years ago
3 The volume of a gas is 50.0 mL at 20.0 K. What will be the new
Damm [24]

Answer:

C. 4.00 K

Explanation:

We can solve this using Charles's Law of the ideal gas. The law describes that when the pressure is constant, the volume will be directly proportional to the temperature. Note that the temperature here should only use the Kelvin unit. Before compressed, the volume of the gas is 50ml(V1) and the temperature is 20K (T1). After compressed the volume becomes 10ml(V2). The calculation will be:

V1 / T1= V2 / T2

50ml / 20K = 10ml / T2

T2= 10ml/ 50ml * 20K

T2= 4K

7 0
3 years ago
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