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HACTEHA [7]
3 years ago
8

18.2 mL of a 0.156 M solution of lead(II) nitrate are added to 26.2 mL of a 0.274 M solution of potassium iodide.

Chemistry
1 answer:
kow [346]3 years ago
5 0

The mass of the Pbl2 : 1308.87

<h3>Further explanation</h3>

Given

18.2 mL of a 0.156 M Pb(NO3)2

26.2 mL of a 0.274 M KI

Reaction

Pb(NO3)2 (aq) + 2 KI (aq) - Pbl2 (s) + 2 KNO3

Required

the mass of the Pbl2

Solution

mol Pb(NO3)2 = 18.2 x 0.156 = 2.8392 mlmol

mol KI = 26.2 x 0.274 =7.1788 mol

Limiting reactant Pb(NO3)2(smaller ratio of mol : reaction coeffiecient)

mol Pbl2 based on limiting reactant (Pb(NO3)2)

From equation, mol ratio of Pb(NO3)2 : Pbl2 = 1 : 1, so mol Pbl2=mol Pb(NO3)2=2.8392

Mass Pb(NO3)2 :

\tt mass=mol\times MW\\\\mass=2.8392\times 461\\\\mass=1308.87~grams

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Answer:

1. The steps of the scientific method consist of seven things: Ask a question, get the research needed for a experiment, (What are you trying to find out? What materials do I need to complete the experiment? What are the procedures needed to complete the experiment?) create a hypothesis, (your educated guess on what you think the results of the experiment will be) conduct a experiment (complete multiple different trials preferably three) to test your hypothesis, make observations and record data during the experiment, draw a conclusion, (Was your hypothesis correct? Yes or no and explain why it is wrong or right) present your findings.

2. A hypothesis is a educated guess that the scientist believe will be the answer at the end of the experiment which is why they conduct a experiment in the first place to find viable data that will support their hypothesis.

3.  Predictions consist of theories that will test the hypothesis (educated guess) because they're the reasoning to why you believe your guess is correct.

4. A control group consists of variables that do not go through change during a experiment, things that remain the same.

5. Data can be presented in many different ways in the form of graphs, charts, or a research paper. You find the data by completing multiple different trials in a experiment, to make sure you have valid results to write down in your data research.

6. Remember that your hypothesis is your educated guess at the beginning of the experiment, what YOU thought was going to happen during the experiment and if the data you received during the experiment supports your hypothesis.

7. Your procedures, perhaps you measured a variable wrong, perhaps you used to much, or to little of a variable, all depends on your experiment.

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Hope this helps.

4 0
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What is the balanced form of the chemical equation shown below?
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Answer:

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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
3 years ago
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