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Hoochie [10]
3 years ago
5

Can you guys help me.

Chemistry
1 answer:
Alisiya [41]3 years ago
5 0
<span>1. Formula 2. Reactants 3. Atoms & molecules 4. Destroyed or created 5. Balanced 6. Mass number 7. 2Al(s) + 3CuCl2(aq) = 2AlCl3(aq) + 3Cu(s)</span>
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NAME THESE ALCOHOLS. CH3CH2CH2CH2OH AND CH3CH(OH)CH3
ale4655 [162]
"CH3CH2CH2CH2OH " is known by the name of "n-butanol" and "CH3CH(OH)CH3" is known by the name of "<span>Isopropyl alcohol". These two given products are basically alcohols. I hope that this is the answer that you were looking for and the answer has actually come to your desired help. Thanks for joining brainly and getting your questions solved.</span>
3 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
3 years ago
What would be the oxidation number of the atoms in these compounds?
sashaice [31]

Answer:

Oxidation state] is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules:

The oxidation state of a free element (uncombined element) is zero for a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion.

Hydrogen has an oxidation state of 1 and oxygen has an oxidation state of −2 when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of −1 in hydrides of active metals, e.g. LiH, and oxygen has an oxidation state of −1 in peroxides, e.g. H2O2 the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion.

The same is written in my textbook. But how am I supposed to find the ox. number of an atom, which is in compound like K2UO4?

8 0
3 years ago
3. How many moles of HCl are present in 3.70 L of a 2.33 M HCl solution?
Lubov Fominskaja [6]

Answer:

8.6 mol

Explanation:

number of moles = molar concentration x volume in litre

number of moles = 2.33 M x 3.70 L = 8.6 mol

4 0
3 years ago
Adrian is measuring the time it takes to heat 200 mL of water until it boils. He is doing the experiment for the third time. He
faltersainse [42]

Answer:

I think a

explanation

..

6 0
3 years ago
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