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2 years ago

BRAINLIEST IF ANSWERED CORRECTLY WITH EXPLANATION! HELP PLEASE:)

Chemistry

1 answer:

elena-14-01-66 [18.8K]2 years ago

4 0

First question

$Pb(NO_3)_2 + KI \rightarrow PbI_2 + KNO_3$

Balaneced:

$Pb(NO_3)_2 + 2KI \rightarrow PbI_2 + 2KNO_3$

Find how many moles potassium iodide it is in 345.0 grams

$n_{KI} = \frac{345.0 \ g}{166.0 \ g/mol} = 2.078 \ mole$

Find how many moles potassium nitrate there is for 2.078 moles $KI$

$n_{KNO_3} = \frac{2}{2} \cdot n_{KI} = 2.078 \ mole$

Find how many grams potassium nitrate there is in 2.078 moles $KNO_3$

$m_{KNO_3} = 2.078 \ mol \cdot 101.1 \ g/mol = 210.1 \ g$

Second question

$yield \ \% = \frac{experimental}{theoretical} \cdot 100 = \frac{195.3 \ grams}{210.1 \ grams} \cdot 100 = 93.0 \ \%$

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Dominic made the table below to organize his notes about mixtures. A 1-column table. The first column labeled properties of mixt

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2 years ago

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The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abb

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The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abbreviation (w/v) is also common. 262 grams of sucrose are needed to make 655 mL of a 40.0% (w/v) sucrose solution

<h3>Define Solute</h3>

A solute is a material that dissolves in a solution. The amount of solvent present in fluid solutions is greater than the amount of solute. The two most common examples of solutions in daily life are salt and water. Salt is the solute because it dissolves in water.

<h3>forms of ratios for product concentration or yield:-</h3>

w/v:- Weight by volume or weight per volume are the terms used. Any solid compound's concentration in a liquid can be calculated using it. It is measurable in gm/ml.

Weight by weight ratio is referred to as w/w.It is employed to determine the final yield of the compound obtained from the starting compound. as in —mg/—gm.

It provides the real yield of the substance or item.

Volume/volume. It is used to specify a liquid's composition or percent in a liquid compound.

using w/v we can calculate the weight of sucrose:-

40.0% means 40 g sucrose/ 100 g solution

40.0g sucrose x (655/100)=grams of sucrose

262 grams of sucrose are needed to make 655 mL of a 40.0% (w/v) sucrose solution.

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1 year ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

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2 years ago

1. An electric power plant uses energy from burning coal to generate steam at 450 °C. The plant is cooled by 20 °C water

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Answer:

Explanation:

Efficiency of the electric power plant is $e=1-\frac{T_{2}}{T_{1}}$