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4 years ago

If the enthalpy change of a reaction in a flask is δh = +67 kj, what can be said about the reaction?

Chemistry

2 answers:

san4es73 [151]4 years ago

5 0

The enthalpy change of the reaction indicates that it is an endothermic process.

FURTHER EXPLANATION

Enthalpy (ΔH) is the amount of heat absorbed or released in a reaction. It is based on the amount of energy needed to break the bonds and the energy released during bond formation. Enthalpy change is the difference in the enthalpy of the reactants and the products. The positive or negative sign for an enthalpy value indicates the direction of the heat flow: a positive ΔH indicates that the reaction is endothermic while a negative value for ΔH means that the reaction is exothermic.

Endothermic Reactions

Endothermic reactions are reactions that absorb heat from the surroundings to the system. This is the case when more energy is absorbed to break the bonds than is released to form the bonds. Endothermic reactions can be identified in the lab by observing if the reaction vessel becomes cooler as the reaction proceeds.

Exothermic Reactions

When the amount of energy released during bond formation is greater than the amount of energy absorbed during bond breaking, a net release of energy to the surroundings takes place and the reaction is exothermic. Exothermic reactions can be identified when the reaction vessel becomes hot as the reaction progresses.

LEARN MORE

Heat capacity brainly.com/question/12976104

Equilibrium brainly.com/question/12983923

Gibbs Free Energy brainly.com/question/12979420

Keywords: Endothermic, Exothermic, Enthalpy

ziro4ka [17]4 years ago

3 0

Answer : Normally in any chemical reaction, if the enthalpy change i.e. ΔH is positive which means it is greater than zero then it can be called as an Endothermic Reaction.

Whereas, the system under study is absorbing heat that is produced during the reaction. So if ΔH is found to be positive then it can be called as endothermic reaction.

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How many formula units make up 24.2 g of magnesium chloride (MgCl2)? Help!!

NNADVOKAT [17]

Answer:

Approximately $1.53 \times 10^{23}$ formula units ( $0.254\; \rm mol$ ).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium ( $\rm Mg$ ) and chlorine ( $\rm Cl$ ):

$\rm Mg$ : $24.305$ .
$\rm Cl$ : $35.45$ .

In other words, the mass of $1\; \rm mol$ of $\rm Mg$ atoms would be (approximately) $24.305\; \rm g$ .

Likewise, the mass of $1\; \rm mol$ of $\rm Cl$ atoms would be approximately $35.45\; \rm g$ .

One formula unit of the ionic compound $\rm MgCl_{2}$ includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of $1\; \rm mol$ of the formula units of this compound.

The formula $\rm MgCl_{2}$ includes one $\rm Mg$ atom and two $\rm Cl$ atoms.

Hence, every formula unit of $\rm MgCl_{2} \!$ would include the same number of atoms: one $\rm Mg\!$ atom and two $\rm Cl\!$ atoms. There would be $1\; \rm mol$ of $\rm Mg$ atoms and $2\; \rm mol$ of $\rm Cl$ atoms in $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units.

Thus, the mass of $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units would be equal to the mass of $1\; \rm mol$ of $\rm Mg$ atoms plus the mass of $2\; \rm mol$ of $\rm Cl$ atoms. (The mass of $1\; \rm mol\!\!$ of each atom could be found from the relative atomic mass of each element.)

$\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the formula mass of $\rm MgCl_{2}$ is $95.205\; \rm g \cdot mol^{-1}$ .

Therefore, the number of formula units in $m = 24.2\; \rm g$ of $\rm MgCl_{2}$ would be:

$\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}$ .

Multiple $n$ by Avogadro's Number $N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}$ to estimate the number of formula units in $0.254\; \rm mol$ :

$\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}$ .

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3 years ago

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zavuch27 [327]

The answer is 7 my guy

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AVprozaik [17]

Answer:

Explanation:

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3 years ago

Calculate the pH and pOH of 500.0 mL of a phosphate solution that is 0.285 M HPO42– and 0.285 M PO43–. (Ka for HPO42- = 4.2x10-1

jekas [21]

Answer: when concentrations of acid and base are same, pH = pKa

PH = 12.38 pOH = 1.62

Explanation: pKa= -log(Ka)= 12.38. PH + pOH = 14.00

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marysya [2.9K]

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