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ki77a [65]
3 years ago
8

How old is the idea that matter is composed of atoms?

Chemistry
1 answer:
In-s [12.5K]3 years ago
6 0
The atom theory has been around since the 19th centry 
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When the [H+] in a solution is 1.7 × 10−9 M, what is the pOH ?
Marysya12 [62]
<em>[H⁺] = 1,7×10⁻⁹  mol/dm³</em>

pH = -log[H⁺]
pH = -log[<span>1,7×10⁻⁹]
<u>pH = 8,77</u>

pH + pOH = 14
pOH = 14-8,77
<u>pOH = 5,23

</u>:)
</span>
7 0
3 years ago
Read 2 more answers
In 1993 the Minnesota Department of Health set a health risk limit for chloroform in groundwater of 60.0 g/L Suppose an analytic
tangare [24]

Answer:

4.74 × 10³ mg

Explanation:

Given data

  • Health risk limit for chloroform in groundwater: 60.0 g/L
  • Volume of the sample of groundwater: 79.0 mL = 79.0 × 10⁻³ L

The maximum mass of chloroform that there could be in the sample of groundwater to meet the standards are:

79.0 × 10⁻³ L × 60.0 g/L = 4.74 g

1 gram is equal to 10³ milligrams. Then,

4.74 g × (10³ mg/1 g) = 4.74 × 10³ mg

6 0
3 years ago
Which elements would have properties similar to those of boron (B) ? Se , Li , and Be Al , Ga , and In C, N, and O Na , Mg , and
n200080 [17]

Answer:

option B= Al, Ga and In

Explanation:

Properties of Boron group:

  1. These elements are belongs to boron group which is thirteen group of periodic table. There are six elements are in this group boron, aluminium, gallium, indium, thallium and nihonium.
  2. They are also called p-block elements because their valance electrons are present in p subshell.
  3. All these elements have three valance electrons.
  4. Boron is metalloid while other elements are metals.
  5. Their oxidation state is +3 because of tendency to lose three valance electrons and create positive charge.
  6. Their atomic sizes increases down the group with increase of atomic number.
  7. Their ionization energies decreases down the group because of increase of atomic radius and ease of removing of electrons.

8 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
3 years ago
How many Faradays of electricity are carried by 1.505×10^23 electronsk​
aksik [14]

Mole of electron required by 1.505 *10^{23} mole is 2.5* 10  ^{-1}

  • Faraday law expressed how the  change that is been being produced by a current at an electrode-electrolyte interface is related and  proportional to the quantity of electricity  that is been used.

  • There is one mole of electron required for 1 Faraday of electricity.

  • Avogadro constant is 6.02*10^{23}

  • Mole of electron can be calculated by dividing the number of electron by avogadro's constant.

=\frac{1.505*10^{23} }{6.02*10^{23} }

= 2.5* 10^{-1}  Faraday  of electricity

Therefore, it requires  2.5*10^{-1} Faraday of electricity for the 1.505 *  10^{23}  mole.

Learn more at: brainly.com/question/1640558?referrer=searchResults

3 0
2 years ago
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