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Yanka [14]
3 years ago
10

PLEASE HELP!!!!!

Chemistry
1 answer:
Blababa [14]3 years ago
4 0
It would be water. Since the body is 60-75% water, it is important in chemical reactions within the body.
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What property of an element does the group number identify in a numbering system that uses "A" and "B"?
VMariaS [17]
<span>The number of valence electrons. :) have a good day </span>
8 0
3 years ago
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How do you calculate the number of valence electrons in an atom?
Masja [62]
By calculating it's number of GROUP........
5 0
2 years ago
Determine the a) energy (in eV) and b) wavelength (in cm) corresponding to blue light of frequency 670 THz
Lesechka [4]

Answer :

(a) The energy of blue light (in eV) is 2.77 eV

(b) The wavelength of blue light is 4\times 10^{-5}cm

Explanation:

The relation between the energy and frequency is:  

Energy=h\times Frequency

where,

h = Plank's constant = 6.626\times 10^{-34}J.s

Given :

Frequency = 670THz=670\times 10^{12}s^{-1}

Conversion used :

1THz=10^{12}Hz\\1Hz=1s^{-1}\\1THz=10^{12}s^{-1}

So,  

Energy=(6.626\times 10^{-34}J.s)\times (670\times 10^{12}s^{-1})

Energy=4.44\times 10^{-19}J

Also,  

1J=6.24\times 10^{18}eV

So,  

Energy=(4.44\times 10^{-19})\times (6.24\times 10^{18}eV)

Energy=2.77eV

The energy of blue light (in eV) is 2.77 eV

The relation between frequency and wavelength is shown below as:

Frequency=\frac{c}{Wavelength}

Where,

c = the speed of light = 3\times 10^8m/s

Frequency = 670\times 10^{12}s^{-1}

So, Wavelength is:

670\times 10^{12}s^{-1}=\frac{3\times 10^8m/s}{Wavelength}

Wavelength=\frac{3\times 10^8m/s}{670\times 10^{12}s^{-1}}=4\times 10^{-7}m=4\times 10^{-5}cm

Conversion used : 1m=100cm

The wavelength of blue light is 4\times 10^{-5}cm

7 0
3 years ago
A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
2 years ago
how much hydrogen gas is necessary to exert a pressure of 1.4 ATM at 430k if occupying a volume of 15.1 l
lora16 [44]

Answer:

5

Explanation:

7 0
3 years ago
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