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tatyana61 [14]
2 years ago
14

How much heat is given off when you freeze 95.7 g of water?

Chemistry
2 answers:
vazorg [7]2 years ago
7 0

Answer:

Hey kimalonzo59!

· Calculate specific heat as c = Q / (mΔT). In our example, it will be equal to c = -63,000 J / (5 kg * -3 K) = 4,200 J/ (kg·K). This is the typical heat capacity of water.

<em>I hope this Helps, Have an Awesome Day!</em>

<em>~ Chloe Marcus ❤</em>

forsale [732]2 years ago
6 0

im not too sure but the answer above helps a lot

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How long does it take the sun to melt a block of ice at 0∘c with a flat horizontal area 1.0 m2 and thickness 1.8 cm ? assume tha
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This is a problem involving heat transfer through radiation. The solution to this problem would be to use the formula for heat flux.

ΔQ/Δt = (1000 W/m²)∈Acosθ

A is the total surface area:
A = (1 m²) + 4(1.8 cm)(1m/100 cm)(√(1 m²))
A = 1.072 m²

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ΔQ = mΔHfus
Let's find its mass knowing that the density of ice is 916.7 kg/m³.
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3 years ago
Which of these statements is true about the temperature of a substance?
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(c)

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Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

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Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

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