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muminat
2 years ago
8

A student is trying to develop a model of an instant heat pack. What is the essential characteristic of the chemical reaction th

at the student should
use to develop the heat pack?
O A The standard enthalpy of reaction should be zero.
о В.
The standard enthalpy of reaction should be negative.
OC. The enthalpy of formation of reactants should be positive.
D. The enthalpy of formation of products should be negative.
Chemistry
1 answer:
Lana71 [14]2 years ago
3 0

The standard enthalpy of reaction should be negative.

<h3>What is enthalpy?</h3>

A thermodynamic quantity equivalent to the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume.

Inside the heat pack are two chemicals that get mixed when you smush them together. As they mix, some weak bonds are broken, which takes a little bit of energy. But new, stronger bonds form which release energy. Releasing that energy causes the surroundings to heat up.

Hence, option B is correct.

Learn more about enthalpy here:

brainly.com/question/13775366

#SPJ1

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A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
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egoroff_w [7]

Answer:

Explanation:

Reaction given

6 H⁺ + 2 MnO₄⁻ + 5 (COOH)₂ = 10CO₂ +8H₂O + 2 Mn⁺²

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=  x - 4 x 2 = -1

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Oxidation no of Mn in Mn⁺² =  +2

So its oxidation no is decreased from + 7 to + 2  . Hence it is reduced.

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