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2 years ago

Which of the following steps will slow down the reaction between coal and oxygen?

Chemistry

2 answers:

Kay [80]2 years ago

7 0

Answer: I believe it it decrease temp of oxygen

Explanation:

irakobra [83]2 years ago

5 0

Answer:

decrease temperature of the oxygen

Explanation:

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Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as hept

vlada-n [284]

Answer:

Vegetable oil will dissolve in heptane

Isopropyl alcohol will dissolve in water

Potassium Bromide will dissolve in water

Explanation:

In chemistry, like dissolves like. This implies that polar substances will dissolve in polar solvents and nonpolar substances will dissolve in nonpolar solvents. This is so because, dissolution of a solute in a solvent involves adequate intermolecular interaction between solute and solvent which isn't possible between a polar and a nonpolar substance.

5 0

3 years ago

You are trying to determine the specific heat of a metal. You heat the 97 g piece of metal to 100 °C and place it in a calorimet

Brut [27]

Answer : The specific heat of the metal is, $0.658J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of metal = 97 g

$m_2$ = mass of water = 122 g

$T_f$ = final temperature of mixture = $28.9^oC$

$T_1$ = initial temperature of metal = $100^oC$

$T_2$ = initial temperature of water = $20.0^oC$

Now put all the given values in the above formula, we get

$97g\times c_1\times (28.9-100)^oC=-122g\times 4.18J/g^oC\times (28.9-20.0)^oC$

$c_1=0.658J/g^oC$

Therefore, the specific heat of the metal is, $0.658J/g^oC$

8 0

3 years ago

90 POINTSSSS!!!

Katarina [22]

Answer:

$\large \boxed{\text{-2043.96 kJ/mol}}$

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn: C₃H₈(g) + O₂(g) ⟶ CO₂(g) + H₂O(g)

Balanced eqn: C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}$

(b)Total enthalpies 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)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\= \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}$