In metallic bonds, the mobile electrons surrounding the positive ions are called a(n)

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Answer 1
Answer: It is called, Noble gas

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Give an example of experimental bias
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Chemistry work here. Please help as soon as possible. I have allot of questions that needs to be answered. Can someone do it for me?

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Answer:

I dont know about the yield but I think mno2 is the limiting reactant

A sample containing only carbon, hydrogen, and silicon is subjected to elemental analysis. After complete combustion, a 0.7020 g sample of the compound yields 1.4 g of CO2, 0.86 g of H2O, and 0.478 g of SiO2. What is the empirical formula of the compound?

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Answer: The empirical formula of compound is C_4H_(12)Si.

Explanation:

Mass of Sample= 0.702 g

Mass of CO_2 = 1.4 g

Mass of H_2O = 0.86 g

Mass of SiO_2 = 0.478 g  

First we have to calculate moles ofCO_2, H_2O and SiO_2 formed.

1. Moles of CO_2=(1.4g)/(44g/mol)=0.032mol

Now , Moles of carbon == Moles of CO_2 = 0.032

2.  Moles of H_2O=(0.86g)/(18g/mol)=0.048mol​​​

Now , Moles of hydrogen = 2* Moles of H_2O =2* 0.048=0.096mol

3.  Moles of SiO_2=(0.478g)/(60g/mol)=0.008 mol

Now , Moles of silicon = Moles of SiO_2 = 0.008 moles

Therefore, the ratio of number of moles of C : H : Si is  = 0.032 : 0.096 : 0.008

For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C= (0.032)/(0.008)=4

For H =(0.096)/(0.008)=12

For Si=(0.008)/(0.008)=1

Thus, C: H: Si = 4 : 12 : 1

The simplest ratio represent empirical formula.

Hence, the empirical formula of compound is C_4H_(12)Si.

What are two causes of soil loss?​

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Answer: The agents of soil erosion are the same as of other types of erosion for example water, ice, wind, and gravity. Soil erosion is more likely where the ground has been disturbed by agriculture, grazing animals, logging, mining, construction, and recreational activities.Basically what I mean is some causes of solid loss is mining, construction

A chemist titrates 90.0 mL of a 0.5870 M acetic acid (HCH, CO) solution with 0.4794M NaOH solution at 25 °C. Calculate the pH at equivalence. The p Kg of acetic acid is 4.76. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added

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Answer:

9.09

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

A balloon has a volume of 7.00 liters at a pressure of 740 mm Hg. If the temperature remains constant, at what pressure will the volume decrease to 2.00 liters?a. 749 mm Hg
b. 52.9 mm Hg
c. 211 mm Hg
d. 2590 mm Hg

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Answer:

D

Explanation:

P1 x V1 = P2 x V2

7.00(740) = x(2.00)

2590 mm Hg

took me a while because i dont know this, hope its right good luck

A pharmacist wishes to strengthen a mixture from 10%alcohol to 30% alcohol. How much pure alcohol should be added to 7 liters of the 10% mixture?

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Answer:

2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.

Explanation:

Suppose x is the number of litres added to the 10% mixture than the quantity of new mixture is given as below

  • n_(old)=7 litres
  • n_(new)=7+x litres

Also the quantity of alcohol is given as

  • q_(old)=10 \% \, of \, 7 \, litres =0.7
  • q_(added)=x
  • q_(new)= 30 \% \,of \,new\, quantity = 0.3(7+x)

Now the equation is as

                                  q_(old)+q_(added)=q_(new)\n0.7+x=0.3(7+x)\n0.7+x=2.1+0.3x\nx-0.3x=2.1-0.7\n0.7x=1.4\nx=2 \, litres

So 2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.