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lina2011 [118]
1 year ago
8

Which of these statements is not necessarily true for two objects in thermal equilibrium?

Chemistry
1 answer:
Dmitrij [34]1 year ago
5 0

1. Answer: C. The objects' temperatures have both changed by the same amount.

Explanation:

An object is said to be in thermal equilibrium when the objects have attained same temperature. Heat transfer from hotter object to colder one in contact takes place until the temperature of the two are equal. It is not necessary that the temperature of both the objects changes by same amount. After attainment of thermal equilibrium, the temperature of the objects stop changing and the tiny particles of the object move at the same rate.

Hence, the objects' temperatures have both changed by the same amount. is not necessarily true for two objects in thermal equilibrium.

2. Answer: C. Objects are made of tiny particles, and their motion depends on the temperature.

Explanation:

Kinetic theory of heat states that the kinetic energy of constituent particles determine the temperature of the object. The statement that best explains this is Objects are made of tiny particles, and their motion depends on the temperature.

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CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)
REY [17]
Your answer is right.

Important elements to consider:

- to use the balanced equation (which you did)
- divide the masses of each compound by the correspondant molar masses (which you did)
- compare the theoretical proportions with the current proportions

Theoretical: 2 mol of Na OH : 1 mol of CuSO4
Then 4 mol of NaOH need 2 mol of CUSO4.

Given that you have more than 2 mol of of CUSO4 you have plenty of it and the NaOH will consume first, being this the limiting reagent.

6 0
2 years ago
The density of whole blood is 1.05 g/ml. a typical adult has between 4.7 and 5.5 l of whole blood. what is the mass in pounds of
Kisachek [45]
<span>1.05 g/ml * 1000 ml = 1050g/l because of 1g/ml = 1 kg/l so, a/q mass of 4.7 l of whole blood in pound = 4.7 * 1050 = 4935 g so in pound 4935g = 10.87981p</span>
5 0
2 years ago
There are many important laws and theories in science. Which of the following is a law of science?
Alenkasestr [34]

Answer: the answer is C

Explanation:

4 0
2 years ago
Read 2 more answers
5.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the
cupoosta [38]

Answer: Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

Explanation:

To calculate the entalpy, we use the equation:

q=mc\Delta T

where,

q = heat absorbed by water = ?

m = mass of water = {\text {volume of water}}\times {\text {density of water}}=100.0ml\times 1.00g/ml=100.0g

c = heat capacity of water = 4.186 J/g°C

\Delta T= change in temperature = 6.70^0C

q=100.0g\times 4.184J/g^0C\times 6.70^0C=2803.3J=2.8033kJ

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

The heat absorbed by water will be equal to heat released by MgSO_4

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 5.11 g

Molar mass  = 120 g/mol

Putting values in above equation, we get:

\text{Moles of }MgSO_4=\frac{5.11g}{120g/mol}=0.042mol

0.042 moles of MgSO_4 releases = 2.8033 kJ

1 mole of MgSO_4 releases = \frac{2.8033 kJ}{0.042}\times 1=66.7kJ

Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

3 0
2 years ago
The enthalpy change for the explosion of ammonium nitrate with fuel oil is –7198 kJ for every 3 moles of NH4NO3. What is the ent
anastassius [24]

Answer:

−2399.33 kJ

Explanation:

If NH₄NO₃ reacts with fuel oil to give a ΔH of -7198 for every 3 moles of NH₄NO₃

What is the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction

∴ For every 1 mole, we will have \frac{1}{3} of the total enthaply of the 3 moles

so, to determine the 1 mole; we have:

\frac{1}{3}*(-7198kJ)

= −2399.33 kJ

∴ the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction = −2399.33 kJ

7 0
2 years ago
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