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2 years ago

1. How many joules of heat are necessary to melt 500 g of ice at its freezing point?

Chemistry

1 answer:

ololo11 [35]2 years ago

7 0

Answer: To melt 500 g of ice, 4000 joule of heat is required.

Explanation:

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Complete this equation for the dissociation of NH4NO3(aq).

Delvig [45]

Dissociation
NH₄NO₃(aq) = NH₄⁺(aq) + NO₃⁻(aq)

the hydrolysis of the cation
NH₄⁺(aq) + H₂O(l) = NH₃(aq) + H₃O⁺(aq)
pH<span><7</span>

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3 years ago

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If 1.16 L of water is initially at 24.2 ∘C, what will its temperature be after absorption of 9.4×10−2 kWh of heat?

vitfil [10]

Answer:

The temperature will be 93.92 °C

Explanation:

To explain this we will use following equation: also Q = ∆U + W known as the NON-FLOW ENERGY EQUATION (N.F.E.E.)

With Q = heat added to the system

with ∆U = change in internal energy

⇒∆U = ( m )( Cv )( T2 - T1 )

With W = work done by the system

⇒For this situation W = 0 because there isn't work done

So we get: ∆U = ( m )( Cv )( T2 - T1 ) = Q

To find the temperature, we have to isolate T in the equation:

(T2-T1) = Q / (m)(Cv)

⇒ Since we know that m = density * volume we can calculate the mass of water.

mass = 1000g/L * 1.16 L = 1160g

Cv = heat capacity ⇒ water has a heat capacity of 4.184 J/g °C

We know the absorption of heat is 9.4x 10^-2 kWh but to know how many joule this is we should convert ( 1 joule = 3.6 x 10^6 kWh)

⇒Q = ( 0.094 kWh ) ( 3.6 x 10^6 J / kWh ) = 0.3384 x 10^6 J

For the temperature we get then: T2 -T1 = Q / (m)(Cv)

T2 - T1 = 0.3384 x 10^6 J / (( 1160g)*(4.184 J/g °C)) = 69.72 ° C

T2 = ( T2 - T1 ) + T1 ⇒ 69.72 + 24.2 = 93.92 °C

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3 years ago

What organ is a muscle and expands and contracts to allow lungs to take in air

jeka57 [31]

Answer:

Your diaphragm contracts and expands

4 0

3 years ago

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Use the given heats of formation to calculate the enthalpy change for this reaction. B2O3(g) + 3COCl2(g) →2BCl3(g) + 3CO2(g) ΔHo

UkoKoshka [18]

Answer:

the enthalpy change for this reaction is -57.7 kJ/mol

Explanation:

Given:

HB₂O₃ = -1272.8 kJ/mol

HCOCl₂ = -218.8 kJ/mol

HBCl₃ = -403.8 kJ/mol

HCO₂ = -393.5 kJ/mol

Those are all standard enthalpies

Question: Calculate the enthalpy change for this reaction, ΔHreaction = ?

The enthalpy of the reaction is calculated using the standard enthalpies of formation of both products and reagents. To understand better, the reaction is as follows

B₂O₃ + 3COCl₂ → 2BCl₃ + 3CO₂

Where the compounds on the left are the reactants and the compounds on the right are the products

ΔHreaction = ∑ΔHproducts - ∑ΔHreactants

$delta(H)_{reaction} =((2*(-403.2)+(3*(-393.5))-((1*(-1272.8)+(3*(-218.8))=-57.7kJ/mol$

Please be careful with the signs.

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3 years ago

Name the following compound:

postnew [5]

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3 years ago