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2 years ago

Determine the oxidation state of each of the following species. Pb2+

Chemistry

1 answer:

mafiozo [28]2 years ago

6 0

Answer:

Pb oxidation number is +2

C in CH4 is -4. H is +1. 4H + 1C = 0. ; 4(+1) + C = 0 ; C = -4

O is usually -2. 3O = 3(-2) = -6. ; 2Fe =. +6 ; each Fe is +3

2Ag = +2 since O = -2 ; each Ag = +1

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Please somebody help!!!!!!!!!!!!!

natta225 [31]

Answer:

1.Materials

2. Hypothesis

3.Question

4.Procedure

5.Further questions

6.Results and Data

7.Conclusion

Explanation:

4 0

2 years ago

If a lab requires each a lab group (3 students) to have 25 ml of a solution and it takes 15 grams of AgNO₃ cuprous nitrate, to m

pochemuha

Answer:

0.375 grams are needed to make 25 mL solution.

Explanation:

Mass of $AgNO_3$ cuprous nitrate required to make 1 l of solution = 15 g.

1 L = 1000 mL

Mass of $AgNO_3$ cuprous nitrate required to make 1000 mL of solution = 15 g

Mass of $AgNO_3$ cuprous nitrate required to make 1 mL of solution:

$=\frac{15}{1000} g$

Mass of $AgNO_3$ cuprous nitrate required to make 25 mL of solution:

$=\frac{15}{1000} \times 25 g=0.375 g$

0.375 grams are needed to make 25 mL solution.

4 0

2 years ago

Can someone help me with this question please?

sineoko [7]

Answer:

Explanation:

4 0

1 year ago

Read 2 more answers

Given the following information: Mass of proton = 1.00728 amu Mass of neutron = 1.00866 amu Mass of electron = 5.486 × 10^-4 amu

lesya692 [45]

Answer: The nuclear binding energy of the given element is $2.938\times 10^{12}J/mol$

Explanation:

For the given element $_3^6\textrm{Li}$

Number of protons = 3

Number of neutrons = (6 - 3) = 3

We are given:

$m_p=1.00728amu\\m_n=1.00866amu\\A=6.015126amu$

M = mass of nucleus = $(n_p\times m_p)+(n_n\times m_n)$

$M=[(3\times 1.00728)+(3\times 1.00866)]=6.04782amu$

Calculating mass defect of the nucleus:

$\Delta m=M-A\\\Delta m=[6.04782-6.015126)]=0.032694amu=0.032694g/mol$

Converting this quantity into kg/mol, we use the conversion factor:

1 kg = 1000 g

So, $0.032694g/mol=0.032694\times 10^{-3}kg/mol$

To calculate the nuclear binding energy, we use Einstein equation, which is:

$E=\Delta mc^2$

where,

E = Nuclear binding energy = ? J/mol

$\Delta m$ = Mass defect = $0.032694\times 10^{-3}kg/mol$

c = Speed of light = $2.9979\times 10^8m/s$

Putting values in above equation, we get:

$E=0.032694\times 10^{-3}kg/mol\times (2.9979\times 10^8m/s)^2\\\\E=2.938\times 10^{12}J/mol$

Hence, the nuclear binding energy of the given element is $2.938\times 10^{12}J/mol$

3 0

2 years ago

Calculate the mass in grams of 3. 29 x 104 mol of lithium Help please?

Sedaia [141]

Answer 

no. of moles =

 $\frac{mass \: {g} \: }{ \: molecular \: wight}$ 

so, mass (g)= no. of moles × molecular wight

Explanation:

Li molecular wight equal to 6.941

mass (g)=3.29×104 mol ×6.941 =2374.93g of lithium 

3 0

2 years ago