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2 years ago

12

State the difference that may be observed between the solids obtain by evaporating a solution of a solid to dryness or letting i

t the solution stand until the solid crystallizes.

1 answer:

tresset_1 [31]2 years ago

3 0

Answer:

It well melt.While theother one will dry

Explanation:

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When a substance is entering a phase change, the gain or loss of heat is a result of:______.

nydimaria [60]

When a substance is entering a phase change, the gain or loss of heat is a result of energy gained or lost in forming or breaking intermolecular interaction.

The constant temperatures occur when a substance is undergoing a phase transition. If heat is removed from a substance , such as in freezing and condensation , then the process is exothermic . In this instance , heat is decreasing the speed of the molecules causing then move slower.

Example : liquid to solid and gas to liquid .

These changes release heat to the surrounding.

To learn more about phase change,

brainly.com/question/12390797

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4 0

2 years ago

Which one of the following statements is not true concerning 2.00 L of a 0.100 M solution of Ca3(PO4)2?

Len [333]

<u>Answer:</u> The correct answer is Option B.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .......(1)

<u>For A:</u>

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 2.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{2.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 2.00L)=0.200mol$

Moles of calcium phosphate = 0.200 moles

<u>For B:</u>

1 mole of calcium phosphate contains 3 moles of calcium atoms, 2 moles of phosphate atoms and 8 moles of oxygen atoms.

So, 0.200 moles of calcium phosphate will contain = $(8\times 0.200)=1.6$ moles of oxygen atoms.

Moles of oxygen atoms = 1.6 moles

<u>For C:</u>

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 1.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{1.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 1.00L)=0.100mol$

Moles of calcium ions = $(0.100\times 3)=0.300$ moles

<u>For D:</u>

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 5.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{5.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 5.00L)=0.500mol$

Moles of phosphorus atoms = $(0.500\times 2)=1.00$ moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of atoms

Number of phosphorus atoms in 0.500 moles of calcium phosphate will be = $(1.00\times 6.022\times 10^{23})=6.022\times 10^{23}$

<u>For E:</u>

1 mole of calcium phosphate contains 3 moles of calcium ions and 2 moles of phosphate ions.

So, 0.200 moles of calcium phosphate will contain = $(3\times 0.200)=0.600$ moles of calcium ions

Moles of calcium ions = 0.600 moles

Hence, the correct answer is Option B.

6 0

3 years ago

A 3.00-kg block of copper at 23.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen

hammer [34]

Answer:

1.2584kg of nitrogen boils.

Explanation:

Consider the energy balance for the overall process. There are not heat or work fluxes to the system, so the total energy keeps the same.

For the explanation, the 1 and 2 subscripts will mean initial and final state, and C and N2 superscripts will mean copper and nitrogen respectively; also, liq and vap will mean liquid and vapor phase respectively.

The overall energy balance for the whole system is:

$U_1=U_2$

The state 1 is just composed by two phases, the solid copper and the liquid nitrogen, so: $U_1=U_1^C+U_1^{N_2}$

The state 2 is, by the other hand, composed by three phases, solid copper, liquid nitrogen and vapor nitrogen, so:

$U_2=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

So, the overall energy balance is:

$U_1^C+U_1^{N_2}=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

Reorganizing,

$U_1^C-U_2^C=U_{2,liq}^{N_2}+U_{2,vap}^{N_2}-U_1^{N_2}$

The left part of the equation can be written in terms of the copper Cp because for solids and liquids Cp≅Cv. The right part of the equation is written in terms of masses and specific internal energy:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_1^{N_2}u_1^{N_2}$

Take in mind that, for the mass balance for nitrogen, $m_1^{N_2}=m_{2,liq}^{N_2}+m_{2,vap}^{N_2}$ ,

So, let's replace $m_1^{N_2}$ in the energy balance:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,liq}^{N_2}u_1^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

So, as you can see, the term $m_{2,liq}^{N_2}u_{2,liq}^{N_2}$ disappear because $u_{2,liq}^{N_2}=u_{1,liq}^{N_2}$ (The specific energy in the liquid is the same because the temperature does not change).

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}(u_{2,vap}^{N_2}-u_1^{N_2})$

The difference $(u_{2,vap}^{N_2}-u_1^{N_2})$ is the latent heat of vaporization because is the specific energy difference between the vapor and the liquid phases, so:

$m_{2,vap}^{N_2}=\frac{m_C*Cp*(T_1^C-T_2^C)}{(u_{2,vap}^{N_2}-u_1^{N_2})}$

$m_{2,vap}^{N_2}=\frac{3kg*0.092\frac{cal}{gC} *(296.15K-77.3K)}{48.0\frac{cal}{g}}\\m_{2,vap}^{N_2}=1.2584kg$

3 0

4 years ago

What process is involved in releasing stored energy from the stored fuel molecules?

Ivanshal [37]

Answer: "exothermic" .
______________________________________________

4 0

3 years ago

Read 2 more answers

How many liters of a 6.0 M solution of acetic acid contain 0.0030 moles

fenix001 [56]

Answer: Least Common Multiplier of a6m, 0.003 moles 0.01elamos

Steps 6m, 0.003 moles

Compute an expression comprised of Factors that appear either in a6m or 0.003 moles

= 0.018elamos

Explanation:

5 0

3 years ago