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Dominik [7]
3 years ago
8

How is hess's law applied in calculating enthalpy

Chemistry
2 answers:
Naddik [55]3 years ago
7 0

Answer: To develop a strategy for the order in which you add the various equations.

Explanation:

In a Hess’s Law problem, you have to develop a strategy for the order in which you add the various equations. hope this helps <3

eimsori [14]3 years ago
3 0

Answer:Hess's law states that the change of enthalpy in a chemical reaction (i.e. the heat of reaction at constant pressure) is independent of the pathway between the initial and final states. ... Hess's law allows the enthalpy change (ΔH) for a reaction to be calculated even when it cannot be measured directly.

Explanation:

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B. It is important that people are not harmed for the sake of science.

Explanation:

Ethical principles stress the need to do good and cause no harm.A researcher is therefore required to;

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4 0
2 years ago
write the steps you would use for making tea.use the words solution,solvent,solute,dissolve,soluble,insoluble,filtrate and resid
poizon [28]
1. Take 100ml of water as solvent and boil it few minutes.
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4 0
3 years ago
A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
A stable atom that has a large nucleus most likely contains
maria [59]

Answer:

3. Equal numbers of  protons and neutrons

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The nucleus becomes unstable if the ratio of protons to neutrons is less than 1:1 or more than 1:1.5.

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3 years ago
The enthalpy of neutralization for the reaction of a strong acid with a strong base is −56 kJ/mol of water produced. How much en
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