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Vlad1618 [11]
3 years ago
12

Plz help i need the answer ASAP If 6.20 cm3 of a 6.75 M (M = mol/dm3) solution are diluted to 85.6 cm3 with water, what is the c

oncentration of the final solution?
A. 6.75 (6.20/85.6)

B. 6.75 (8.56/6.20)

C. 6.75 (6200/85.6)

D. 6.75 (85.6/6200)

a) A
b) B
c) C
d) D
Chemistry
1 answer:
Brilliant_brown [7]3 years ago
6 0

Answer:

Option A. 6.75 (6.20/85.6)

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 6.2cm³

Initial concentration (C1) = 6.75M

Final volume (V2) = 85.6cm³

Final concentration (C2) =.?

The final concentration can be obtained by using the dilution formula as show below:

C1V1 = C2V2

6.75 x 6.2 = C2 x 85.6

Divide both side by 85.6

C2 = (6.75 x 6.2) /85.6

C2 = 6.75 (6.2 / 85.6).

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Compare and contrast alpha beta and gamma radiation
Evgen [1.6K]
There are usually 3 topics used to compare types of radiation: 

Ionising ability
Penetrative power 
Range in air

Ionising ability
Alpha radiation has strong ionising ability, while beta only has moderate ionisation and gamma is very weakly ionising.

Penetrative power 
Alpha particles are weakly penetrating, stopped by paper, while beta particles have stronger penetrating ability, stopped by skin and gamma radiation is very strongly penetrating, stopped only by thick layers of lead.

Range in air
Alpha particles- range of only a few centimetres
Beta - range of up to one meter 
Gamma- infinite range in air.

Hope this helps:)


8 0
3 years ago
Which discovery did J. J. Thomson make that improved upon Dalton's atomic theory?
erik [133]

Answer: Atoms contain tiny, negatively charged electrons

Explanation: Thomson's experiments with cathode ray tubes helped him to discover the electron (which Dalton did not know about). Dalton thought that atoms were indivisible particles, and Thomson's discovery of the electron proved the existence of subatomic particles.

3 0
3 years ago
Are the properties of Rubidium more similar to those of cesium or those of strontium?
fenix001 [56]

More similar to Cesium

Explanation:

The properties of Rubidium are more similar to those of cesium compared to strontium.

Elements in the same group on the periodic table have similar chemical properties.

  • Rubidium and Cesium are located in the first group on the periodic table.
  • Other elements in this group are lithium, sodium, potassium and francium
  • Strontium belongs to the second group on the periodic table.
  • The first group have a ns¹ valence shells electronic configuration.
  • They are all referred to as alkali metals

Learn more:

Sodium brainly.com/question/6324347

Periodic table brainly.com/question/1704778

#learnwithBrainly

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5 0
3 years ago
Complete and balance the following neutralization reaction Ca(OH)2 + H2SO4
Debora [2.8K]

Answer: Ca(OH)2 (aq) + H2SO4 (aq) ---------->  CaSO4(aq) + 2H2O(l)

Explanation:

Since this is a neutralization reaction, the end product would be salt and water. In this equation Calcium will displace hydrogen from the acid because it is more reactive, resulting in the formation of CaSO4 (salt), while the displaced H2 molecule combines with OH molecules to form water.

The equation of the reaction is thus;

Ca(OH)2 (aq) + H2SO4 (aq) ---------->  CaSO4(aq) + H2O(l), in other to balance it, we add ''2'' to the water molecule in the right hand side of the equation.

Balance equation is  

Ca(OH)2 (aq) + H2SO4 (aq) ---------->  CaSO4(aq) + 2H2O(l)

6 0
3 years ago
(1) The solubility of Salt AB2(S) IS 5mol/dm^3.
Svetlanka [38]

Answer:

a. Ksp = 4s³

b. 5.53 × 10⁴ mol³/dm⁹

Explanation:

a. Obtain an expression for the solubility product of AB2(S),in terms of s.

AB₂ dissociates to give

AB₂ ⇄ A²⁺ + 2B⁻

Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as

AB₂ ⇄ A²⁺ + 2B⁻

1 : 1 : 2

Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s

So, we have

AB₂ ⇄ A²⁺ + 2B⁻

[s]        [s]    [2s]

So, the solubility product Ksp = [A²⁺][B⁻]²

= (s)(2s)²

= s(4s²)

= 4s³

b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³

Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³

Substituting the value of s into the equation, we have

Ksp = 4s³

= 4(2.4 × 10³ mol/dm³)³

= 4(13.824 × 10³ mol³/dm⁹)

= 55.296 × 10³ mol³/dm⁹

= 5.5296 × 10⁴ mol³/dm⁹

≅ 5.53 × 10⁴ mol³/dm⁹

Ksp = 5.53 × 10⁴ mol³/dm⁹

6 0
3 years ago
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