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3 years ago

Used to describe an atom’s ability to chemically bond with another atom

Chemistry

1 answer:

zepelin [54]3 years ago

7 0

Answer:

"Electronegativity" is the atom's ability to chemically bond with another atom.

Explanation:

Electronegativity is the ability of the atom to attract the electrons towards another atom to it, these two atoms are associated through a bond. The phenomena is based on the atom's ionization energy and the electron affinity. The formation of the chemical bond takes place when the atoms share the electrons they have in outer shells. In other words we can say two atoms fuse together to form chemical bond.

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Calculate the mass-to-mass ratio of 25.0 g of salt in 105 g of water. Work must be shown in order to earn credit.

Sveta_85 [38]

Answer:

0.23

Explanation:

It is known that, the mass to mass ratio of the salt to water

= (mass of salt / mass of water)

= (25.0 g / 105.0 g)

= 0.23

So, the answer is 0.23

4 0

3 years ago

How many atoms are here in the following?

Elina [12.6K]

Answer:

A. 6N

B. 4H, 2O

C. 4H, 4N, 12O

D. 2Ca, 4O, 4H

E. 3Ba, 6Cl, 18O

F. 5Fe, 10N, 30O

G. 12Mg, 8P, 32O

H. 4N, 16H, 2S, 8O

I. 12Al, 18Se, 72O

J. 12C, 32H

I am 90% sure this is correct

6 0

3 years ago

Approximately how many moles of chlorine make up 1.79 × 1023 atoms of chlorine?

e-lub [12.9K]

1.2*10^24# atoms of chlorine

Explanation:
Chlorine gas (#Cl_2#) has two atoms of elemental chlorine in a molecule, so:
#1# mol of #Cl_2# have #6*10^23# molecules of #Cl_2#
#1# molecule of #Cl_2# have #2# atoms per molucule
Then #2*6*10^23 = 1.2*10^24# atoms of chlorine in a mol of chlorine gas

6 0

3 years ago

Suppose NAD is unavailable because NADH cannot be oxidized due to a mutation in the NADH dehydrogenase (Complex I). If FAD could

Dovator [93]

Answer:

FAD substitution will produce 28 ATP instead of 36.

Explanation:

NAD and FAD are coenzymes involved in reversible oxidation and reduction reactions. These compounds are also known as electron carriers. However NADH produce 3 electrons in electron transport chain and FADH2 produce 2 electron beacuase it transfer the electrons to second complex in ETC.

Normal prduction of ATP from glucose;

2 cytoplasmic NADH formed in glycolysis Each yields 2 ATP +4

2 NADH formed in the oxidation of pyruvate Each yields 3 ATP +6

2 FADH2 formed in the citric acid cycle Each yields 2 ATP +4

6 NADH formed in the citric acid cycle Each yields 3 ATP +18

2 ATP from glycolysis +2

2 ATP from citric acid cycle +2

Net yield ATP +36

C6H12O6 + 6 CO2 + 36 ADP + 36 Pi ⇒6 CO2 + 6 H2O + 36 ATP

If we replace the NAD with FAD the total ATP production would be.

2 cytoplasmic FADH2 formed in glycolysis Each yields 2 ATP +4

2 FADH2 formed in the oxidation of pyruvate Each yields 3 ATP +4

2 FADH2 formed in the citric acid cycle Each yields 2 ATP +4

6 FADH2 formed in the citric acid cycle Each yields 2 ATP +12

2 ATP from glycolysis

2 ATP from citric acid cycle +2

<u>Net yield ATP +28</u>

C6H12O6 + 6 CO2 + 28ADP + 28 Pi ⇒6 CO2 + 6 H2O + 28 ATP

7 0

3 years ago

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb

andre [41]

Answer:

$m_{PbI_2}=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

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

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2$

Best regards.

5 0

3 years ago

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