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nevsk [136]
3 years ago
8

Which part of a chemical formula indicates the number of copies of molecules or compounds?

Chemistry
1 answer:
murzikaleks [220]3 years ago
5 0
That would be the super script, which is the number that appears in front of the molecules composition. 
Ex. 3H2O: The 3 is the super script, and shows that there is 3 water molecules
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Consider the balanced equation:
kakasveta [241]

Answer:

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3 years ago
Which metal can replace another metal in a reaction?
deff fn [24]

Answer:

O A. A metal higher on the activity series list will replace one that is

lower.

4 0
3 years ago
What is the freezing point of a solution made with 1.31 mol of CHCl3 in 530.0 g of CCl4 (Kf =29.8 degrees C/m, Freezing point of
Allisa [31]

73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.

Explanation:

Data given:

number of moles of CHCl3 = 1.31 moles

mass of solvent CHCl3 = 530 grams or 0.53 kg

Kf = 29.8 degrees C/m

freezing point of pure solvent or CCl4 =  -22.9 degrees

freezing point = ?

The formula used to calculate the freezing point of the mixture is

ΔT = iKf.m

m=  molality

molality = \frac{moles of solute}{mass of solvent in kilograms}

putting the value in the equation:

molality= \frac{1.31}{0.53}

             = 2.47 M

Putting the values in freezing point equation

ΔT = 1.31 x 29.8 x 2.47

ΔT = 73.606 degrees

6 0
3 years ago
Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.
o-na [289]

Answer:

  • pKa = 7.46

Explanation:

<u>1) Data:</u>

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

<u>2) Strategy:</u>

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

<u>3) Solution:</u>

a) pH

  • pH = - log [H₃O⁺]

  • 4.64 = - log [H₃O⁺]

  • [H_3O^+]= 10^{-4.64} = 2.29.10^{-5}

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka =  [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0
3 years ago
PLEASE SOMEBODY EXPLAIN THIS :(((
djverab [1.8K]

Answer:

\boxed{ \sf \: R_f  \: value \: of \: sample \: 1 =0.3142}

<h3>Explanation:</h3>

In Analytical Chemistry chromatography is widely used for the separation of samples.

  • In thin layer chromatography, the mixture of components are separated on the basis of their polarity.
  • The solvent solution(mobile phase) that we use are non polar & silica gel( TLC paper made of/stationary phase) are polar.
  • Consider the mixture we have taken consist of two samples having large polar difference.
  • Due to opposite nature of silica gel(polar) & solvent solution (non polar) the movement become easy & due to capillary action solvent solution rise to the top.
  • The mixture of sample we have taken, the sample have less polarity have high peak or they travel more distance than that of more polar sample when they dipped into the solution.

In the given diagram, mixture of 8 samples are separated on the basis of their polarity, the distance travelled by solvent is 35 mm, distance travelled by sample 1 is 11 mm & similarly distance travelled by sample 2,3,4,5,6,7 are 15,31,4,22,25,33 in mm respectively.

Rf Value: Rf value is retention factor which tells about relative absorption of each sample & range of Rf value is 0-1.

Formula to calculate Rf value is

\sf R_f  \: value = \frac{distance \: moved \: by \: sample}{distance \: moved \: by \: solvent}

Now, solving for Rf value of sample 1

<em>Given:</em>

Distance moved by sample 1 = 11 mm

Distance movedby solvent = 35 mm

<em>To find:</em>

Rf value of sample 1 = ?

<em>Solution:</em>

Substituting the given data in above formula,

\small \sf R_f  \: value = \frac{distance \: moved \: by \: sample \: 1}{distance \: moved \: by \: solvent}   \\  \small \sf R_f  \: value =  \cancel\frac{11  \: mm}{35 \:  mm}  = 0.3142

\small \boxed{ \sf \: R_f  \: value \: of \: sample \: 1 =0.3142}

<em><u>Thanks for joining brainly community!</u></em>

4 0
2 years ago
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