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Jlenok [28]
4 years ago
13

Avogadro constant is 6.02 1023 (the number of atoms or molecules per mole). What how much charge is there in 1 mole of electrons

? How much charge is there in 1 mole of hydrogen ions (H+). Remember that hydrogen consists of one electron and one proton, so hydrogen ions are just protons.
Chemistry
1 answer:
cupoosta [38]4 years ago
8 0

Answer:

Charge of 1 mole of electrons = -96488.46 C

Charge of 1 mole of protons/ hydrogen ions = +96488.46 C

Explanation:

Given, Avogadro constant:-

N_a=6.023\times 10^{23}

Thus, charge on 1 mole of electron can be calculated as:-

Charge on 1 electron = -1.602\times 10^{-19}\ C

Charge on 6.023\times 10^{23} electrons = -1.602\times 10^{-19}\times 6.023\times 10^{23}\ C = -96488.46 C

Charge of 1 mole of electrons = -96488.46 C

Charge of electron = Charge on proton (magnitude), Only the charge is different. Electron has negative charge and protons have positive charge.

Charge of 1 mole of protons/ hydrogen ions = +96488.46 C

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Which of the following statements best explains why atoms bond?
Keith_Richards [23]

Answer:

atoms bond to attain neutral configuration or to get noble gas configuration.

Explanation:

atoms always want to be stable by attaining noble gas configuration.

they do so by forming bonds. they also bond to each other to make their outer electrons  stable that is if the atom has an excessive electron than its neutral configuration it can bond to another atom that has one less electron than its neutral configuration by ionic bond.

7 0
3 years ago
Find the density of a block with a length of 8.0 cm, a width of
seraphim [82]

Answer:

<h3>Density = 0.5 g/cm³</h3>

Explanation:

Density of a substance can be found by using the formula

Density =  \frac{mass}{volume}

From the question

mass = 32g

To find the density we must first find the volume of the block

From the question the block is a cuboid

Volume of cuboid = length × width × height

length = 8 cm

width = 4 cm

height = 2 cm

Volume of block = 8 × 4 × 2 = 64 cm³

Substitute the values into the above formula and solve for the density

That's....

Density =  \frac{32}{64}  \\  =  \frac{1}{2}

We have the final answer as

Density = 0.5 g/cm³

The block will float because it's density is less than the density of water which is

1 g/cm³

Hope this helps you

6 0
3 years ago
Read 2 more answers
In standardizing the solution of aqueous sodium hydroxide, a chemist overshoots the end point and adds too much naoh(aq). would
irinina [24]

Answer: Concentration of NaOH calculated will be underestimated.

Explanation:

End point is an observational point , which tells us about the completion of reaction between the titrant (solution in burette) and titre(solution in conical flask) in titration experiment.

In this case , NaOH is titrant whose concentration is unknown.

M_1=\text{molarity of titre} , M_2=\text{molarity of NaOH}

V_1=\text{volume of titre} , V_2=\text{volume of NaOH}

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}....(1)

According to question a chemist overshoots the end point and adds to much of NaOH solution, which means increase in the value of V_2.

Then the value of M_2 in equation (1), will get lowered , which means that the concentration of NaOH was lower than that of the actual value. Hence underestimated concentration of NaOH.




8 0
3 years ago
What is the length of one mole of the typical cell phone that is approximately 15 cm in length?
Salsk061 [2.6K]

i dont know how to remove my answer so ignore this

4 0
3 years ago
Read 2 more answers
Th e molar absorption coeffi cients of tryptophan and tyrosine at 240 nm are 2.00 × 103 dm3 mol−1 cm−1 and 1.12 × 104 dm3 mol−1
nataly862011 [7]

Answer:

5.43x10⁻⁵ dm³mol⁻¹= Concentration of tyrosine

2.58x10⁻⁵ dm³mol⁻¹= Concentration of tryptophan

Explanation:

Lambert-Beer law is:

A = E×C×l

Where A is measured absorbance, E is absorption coefficient, C is concentration of solution and l is optical path length.

For the result at 240nm, it is possible to write:

0.660 = 2.00x10³ dm³mol⁻¹cm⁻¹× Ctry × 1cm + 1.12x10⁴ dm³mol⁻¹cm⁻¹× Ctyr × 1cm <em>(1)</em>

At 280 nm:

0.221 = 5.40x10³ dm³mol⁻¹cm⁻¹× Ctry × 1cm + 1.50x10³ dm³mol⁻¹cm⁻¹× Ctyr × 1cm <em>(2)</em>

<em></em>

Thus:

-2.7× (0.660 = 2.00x10³ × Ctry  + 1.12x10⁴ × Ctyr) =

-1.782 = -5.40x10³Ctry - 3.024x10⁴Ctyr

0.221 = 5.40x10³ × Ctry + 1.50x10³  × Ctyr

-------------------------------------------------------------------

-1.561 = -28740×Ctyr

<em>5.43x10⁻⁵ dm³mol⁻¹= Ctyr</em>

Replacing this value in (1) or (2) it is possible to find Ctry, that is:

<em>2.58x10⁻⁵ dm³mol⁻¹= Ctry</em>

4 0
4 years ago
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