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Usimov [2.4K]
4 years ago
10

How many atoms are in 6.70 moles of sulfur (S)?

Chemistry
1 answer:
Pavel [41]4 years ago
4 0

Answer:

40.444 × 10∧23 atoms

Explanation:

The mole is the unit of amount in chemistry. We know that

1 mole of sulfur = 6.02× 10∧23 atoms

The number 6.02× 10∧23 is called Avogadro's number. It is define as

" Avogadro's number is the number of molecules atoms or ions in one gram molecule of a compound, one gram atom of an element or one gram ions of a substance"

For example:

1 g of hydrogen = 1 mole of hydrogen = 6.02× 10∧23 atoms of hydrogen

18 g of water= 1 mole of water= 6.02× 10∧23 molecules of water

Given data:

moles of sulfur= 6.70

number of atoms= ?

solution:

6.70 × 6.02× 10∧23 = 40.444 × 10∧23 atoms

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8. When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. I
Tamiku [17]

Answer:

The boiling point of the water by adding 2.5 moles of aluminium nitrate will be changed by 4° C.

Explanation:

<u>Step 1:</u> define the formula for an elevation of the boiling point

<u><em>Δ T b  = i *K b *bB</em></u>

⇒with Δ T b = the elevation of the boiling point ( in °C or Kelvin)

⇒with i = van't Hoff  i -factor for the solute. The factor i shows the number of individual particles (typically ions) that are formed by a compound in the solution.

⇒ with Kb = ebullioscopic constant, which depends on the properties of the solvent. (Eventually can be calculated).

⇒ with bB= the molal concentration (molality) of the solute.

Since we have 2 different solutions ( component 1 = sugar and component 2 = aluminium nitrate) of the same molality in the same solvent ( water). We can express this as followed:

Δ T b,2 / Δ T b,1  = (i2 * Kb * bB ) / (i1 * Kb *bB)

⇒after he was simplifying this becomes:

Δ T b,2 / Δ T b,1  = i2 / i1

⇒ Now we can isolate either Δ T b,2 or Δ T b,1:

Δ T b,2  =(i2/i1) x Δ T b,1

<u>Sugar</u>

⇒ is a covalent compound, so it doesn't dissociate in water ⇔ i1=1

<u>Aluminium nitrate</u>

⇒ it's a soluble ionic compound and in solution it will dissociate as the following equation:

Al(NO3)3 (aq) → Al3+(aq)  + 3NO3- (aq)

⇒thus here are 4 particels formed : <u>1</u> Al3+ + <u>3</u>NO3-

i2=4

As given, we also know that the temperature was raised by 1°C after adding sugar ⇒ Δ T b,1 = 1°C

<u>Step 2</u>: insert all the numbers in the formula for boiling point elevation

(i2/i1) x Δ T b,1

(4/1 ) x 1°C = 4°C

⇒ The boiling point of the water by adding 2.5 moles of aluminium nitrate will be changed by 4°C

6 0
3 years ago
Given the reaction 2NaOH + H2SO4 = Na2SO4 + 2H2O, what is the total number of grams of NaOH needed to react completely with 196
ollegr [7]

Answer:

160 g of NaOH

Explanation:

Reaction: 2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

When a base and an acid react, the produce water and a salt. → Neutralization

In this reaction 2 moles of hydroxide need 1 mol of sulfuric acid to react, it is the same to say that 1 mol of sulfuric acid needs 2 moles of NaOH to react. We convert the mass of sulfuric to moles and we propose the rule of three:

196 g / 98 g/mol = 2 moles

1 mol of sulfuric acid needs 2 moles of NaOH to react

Then 2 moles of sulfuric acid will react with (2 . 2) /1 = 4 moles of NaOH

We convert the moles of base to grams → 4 mol . 40 g /1mol = 160 g

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4 years ago
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3 years ago
How many liters of 0.305 M K3PO4 solution are necessary to completely react with 187 mL of 0.0184 M NiCl2 according to the balan
Mashutka [201]

Answer:

6.55 mL of K₃PO₄ are required

Explanation:

We need to propose the reaction, in order to begin:

2K₃PO₄ (aq) + 3NiCl₂(aq) →  Ni₃(PO₄)₂ (s) ↓ + 6KCl (aq)

Molarity = mol/L (Moles of solute that are contained in 1 L of solution.)

M = mol / volume(L).  Let's find out the moles of chloride:

- We first convert the volume from mL to L → 187 mL . 1L / 1000mL = 0.187L

0.0184 M . 0.187L = 0.00344 moles of NiCl₂

Ratio is 3:2. Let's propose this rule of three:

3 moles of chloride react with 2 moles of phosphate

Then, 0.00344 moles of NiCl₂ will react with (0.00344 . 2) /3 = 0.00229 moles of K₃PO₄

M = mol / volume(L) → Volume(L) = mol/M

Volume(L) = 0.00229 mol / 0.350 M = 6.55×10⁻³L

We convert the volume from L to mL → 6.55×10⁻³L . 1000mL /1L = 6.55 mL

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3 years ago
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