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zhannawk [14.2K]
2 years ago
12

Scoring Scheme: 3-3-2-1 Given the starting solution had a concentration of 1.25 M, how many moles of Co[H2O]6Cl2 were available

in the amount of starting solution you used
Chemistry
1 answer:
Pepsi [2]2 years ago
8 0

Answer:

The solution has 0.00994 moles of Co[H2O]6Cl2

Explanation:

Complete question: Given the starting solution had a concentration of 1.25 M, how many moles of Co[H2O]6Cl2 were available in the amount of starting solution you used?

mL= 7.95

Step 1: Data given

Concentration of the starting solution = 1.25 M

Starting solution = Co[H2O]6Cl2

Step 2: Calculate molar mass of Co[H2O]6Cl2

Atomic mass of Co = 58.93 g/mol

Atomic mass of H = 1.01 g/mol

Atomic mass of O = 16.00 g/mol

atomic mass of Cl = 35.45 g/mol

Molar mass of Co[H2O]6Cl2 =

1*58.93 + 12*1.01 + 6*16.00 + 2*35.45 = 237.95 g/mol

Step 3: Calculate number of moles

C = n/v

⇒with C = the Concentration of the starting solution = 1.25 M

⇒with n= the number of moles = to be determined

⇒with v = the volume = 7.95 mL = 0.00795 L

n = C * V

n = 1.25 M * 0.00795 L

n = 0.00994 moles

The solution has 0.00994 moles of Co[H2O]6Cl2

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pashok25 [27]

1. The hydrogen ion concentration [H+] is 2.51×10⁻¹² M

2. The hydroxide ion concentration [OH¯] is 3.98×10¯³ M

<h3>1. How to determine the hydrogen ion concentration, [H+]</h3>

From the question given above, the following data were obtained:

  • pOH = 2.40
  • pH = 14 - pOH = 14 - 2.4 = 11.6
  • Hydrogen ion concentration [H+] =?

pH = –Log [H+]

11.6 = –Log [H+]

Multiply through by –1

–11.6 = Log [H+]

Take the anti-log of –11.6

[H+] = anti-log (–11.6)

[H+] = 2.51×10⁻¹² M

<h3>2. How to determine the hydroxide ion concentration, OH¯</h3>
  • Hydrogen ion concentration [H⁺] = 2.51×10⁻¹² M
  • Hydroxide ion concentration [OH¯] =?

[H⁺] × [OH¯] = 10¯¹⁴

2.51×10⁻¹² × [OH¯] = 10¯¹⁴

Divide both side by 2.51×10⁻¹²

[OH¯] = 10¯¹⁴ / 2.51×10⁻¹²

[OH¯] = 3.98×10¯³ M

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3 0
1 year ago
A balloon occupies 175 cm at 25.0 C. What volume will it occupy at 42.5 °C?
patriot [66]

175 cm³ instead of 175 cm.

Answer:

297.5 cm³

Explanation:

From the question;

Initial volume; V1 = 175 cm³

Initial temperature; T1 = 25°C

Final temperature; T2 = 42.5°C

From Charles law we can find the volume V2 from the equation;

V1/T1 = V2/T2

Making V2 the subject gives;

V2 = V1 × T2/T1

V2 = (175 × 42.5)/25

V2 = 297.5 cm³

4 0
3 years ago
A sample of limestone (calcium carbonate, CaCO3) is heated at 950 K until it is completely converted to calcium oxide (CaO) and
pav-90 [236]

Answer:

Therefore, volume of CO₂ produced in the first step is 9141.404 L

Explanation:

Equations of reactions:

A: CaCO₃(s) ---> CaO(s) + CO₂(g)

B: CaO(l) + H₂O(l) ---> Ca(OH)₂(s)

Molar mass of CaCO₃ = 100 g; molar mass of CaO = 56 g; molar mass of CO₂ = 44 g molar mass of H₂P = 18 g; molar mass of Ca(OH)₂ = 74 g

From equation B, 1 mole of CaO produces 1 mole of Ca(OH)₂

This means that 56 g of CaO produces 74 g of Ca(OH)₂

mass of CaO that produces 8.47 kg or 8470 g of Ca(OH)₂ = 8470 g * 56/74 = 6409.73 g of CaO

Therefore, 6409.73 g of CaO were produced in reaction A

From reaction A, 1 mole of CaCO₃ produces 1 mole CaO and 1 mole of CO₂

Number of moles of CaO in 6409.73 g = 6409.73 g/56 g/mol = 114.46 moles

Therefore, 114.46 moles of CO₂ were produces as well.

Molar volume of gas at STP = 22.4 litres

Volume of CO₂ produced at STP = 114.46 * 22.4 L =2563.904 L

However, the above reaction took place at 950 K and 0.976 atm, therefore volume of CO₂ produced under these conditions are obtained using the general gas equation

Using P₁V₁/T₁ = P₂V₂/T₂

P₁ = 1.0 atm, V₁ = 2563.904 L, T₁ = 273 K, P₂ = 0.976 atm, T₂ = 950 K, V₂ = ?

V₂ = P₁V₁T₂/P₂T₁

V₂ = (1.0 * 2563.904 * 950)/(0.976 * 273)

V₂ = 9141.404 L

Therefore, volume of CO₂ produced in the first step is 9141.404 L

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Use the periodic table to complete each nuclear fusion equation.
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Answer: A:5 B:2 C:15 D:8 E:O

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