4.01 grams of Licl are needed to release this amount of heat
<h3>What does one gram weigh?</h3>
The gram is a unit of mass in the International System of Units (SI) that is equal to one thousandth of a kilogram. It was originally known as the gramme. Gram. This pen cap weighs around one gram. A weight scale like this one may provide a precise mass readout for many different things.
Mass of Licl required to release 5850J of heat to the surroundings.
Let x g Licl required.
Then x g ×∆H = 5850J
x g × 1.46 × 10³ = 5850
x = 5850/ 1.46 × 10³
x = 4.0068
so 4.01g Licl required.
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Answer:
In this phenomenon we talk about ideal gases, that is why in these equations the constant is the number of moles and the constant R, which has a value of 0.082
Explanation:
The complete equation would have to be P x V = n x R x T
where n is the number of moles, and if it is not clarified it is because they remain constant, as the question was worded.
On the other hand, the symbol R refers to the ideal gas constant, which declares that a gas behaves like an ideal gas during the reaction, and its value will always be the same, which is why it is called a constant. The value of R = 0.082.
The ideal gas model assumes that the volume of the molecule is zero and the particles do not interact with each other. Most real gases approach this constant within two significant figures, under pressure and temperature conditions sufficiently far from the liquefaction or sublimation point. The real gas equations of state are, in many cases, corrections to the previous one.
The universal constant of ideal gases is not a fundamental constant (therefore, choosing the temperature scale appropriately and using the number of particles, we can have R = 1, although this system of units is not very practical)
Answer:
Energy is transferred when atoms are rearranged.
Explanation:
Answer:
The answer to your question is the letter C) 5648 kJ/mol
Explanation:
Data
C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11 H₂O
H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol
H° O₂ = 0 kJ / mol
H° CO₂ = -393.5 kJ/mol
H° H₂O = -285.8 kJ/mol
Formula
ΔH° = ∑H° products - ∑H° reactants
Substitution
ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)
ΔH° = -4722 - 3143.8 + 2221.8
Result
ΔH° = -5644 kJ/mol
