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3 years ago

A chemist measures the energy change ?H during the following reaction:

Chemistry

1 answer:

slamgirl [31]3 years ago

5 0

Answer:

Endothermic

It absorbs heat

1.20 × 10³ kJ

Explanation:

Let's consider the following thermochemical equation.

2 H₂O(l) → 2 H₂(g) + O₂(g) ΔH = 572 kJ

Since ΔH > 0, the reaction is endothermic, that is, it absorbs heat when H₂O reacts.

572 kJ are absorbed when 36.03 g of water react. The heat absorbed when 75.8 g of H₂O react is:

75.8 g H₂O × (572 kJ/36.03 g H₂O) = 1.20 × 10³ kJ

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Calculate the mass in grams, in 2.28 mol of H2SO4. Please show your work to receive credit.

solmaris [256]

Answer:

98g

Explanation:

h=1*2=2

s=32

0=16*4=64

64+32+2=98

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2 years ago

What is the Net Force?<br> 2N ><br> 4N -<br> SN

maria [59]

The net force is SN haha

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3 years ago

ΚΕ:<br> m.22<br> 2<br> What is the kinetic energy of 10kg dog running with a velocity of 12m/s?

BigorU [14]

Answer:

The answer is

<h2>720 Joules</h2>

Explanation:

The kinetic energy of a body can be found by using the formula

where

m is the mass

v is the velocity / speed

From the question

mass = 10 kg

velocity = 12 m/s

Substitute the values into the above formula and solve

That's

<h3> $KE = \frac{1}{2} \times 10 \times {12}^{2} \\ = 5 \times 144 \: \:$ </h3>

We have the final answer as

<h3>720 Joules</h3>

Hope this helps you

3 0

3 years ago

A gas mixture at 535.0°C and 109 kPa absolute enters a heat exchanger at a rate of 67.0 m3/hr. The gas leaves the heat exchanger

SVEN [57.7K]

Answer:

the heat rate required to cool down the gas from 535°C until 215°C is -2.5 kW.

Explanation:

assuming ideal gas behaviour:

PV=nRT

therefore

P= 109 Kpa= 1.07575 atm

V= 67 m3/hr = 18.6111 L/s

T= 215 °C = 488 K

R = 0.082 atm L /mol K

n = PV/RT = 109 Kpa = 1.07575 atm * 18.611 L/s /(0.082 atm L/mol K * 488 K)

n= 0.5 mol/s

since the changes in kinetic and potencial energy are negligible, the heat required is equal to the enthalpy change of the gas:

Q= n* Δh = 0.5 mol/s * (- 5 kJ/mol) =2.5 kW

7 0

3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago