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3 years ago

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a vessel contains 12g of an ideal gas at t degree c temperature and 1 atm pressure . when the temperature is increased by 10 deg

ree c at the same volume , the pressure increase by 10% .calculate the volume and intial temperature.

1 answer:

dimaraw [331]3 years ago

6 0

Okiw dokie hope yiy can find th answer

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The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p

Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

$\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}$

(ii) Solve the relationship

Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

56x + 18(7.5 - x) = 267.3

56x + 135 - 18x = 267.3

38x = 132.3

x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

4 0

3 years ago

One phosphorus, three chlorine, one oxygen

4vir4ik [10]

Phosphoryl chloride :

POCl3
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hope this helps!.

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5 0

3 years ago

what is the total amount of energy required to remove the most loosely bound electron from each atom in a mole of gaseous Ca

Temka [501]

Answer:

589.8 kJ/mol

Explanation:

Let's consider the process in which <em>the most loosely bound electron from each atom in a mole atom of gaseous calcium is removed.</em>

Ca(g) + Energy ⇒ Ca¹⁺(g) + 1 e⁻

The energy involved in this process is known as First Ionization Energy. The First Ionization Energy for calcium is 589.8 kJ/mol.

5 0

2 years ago

Complete ionic,net and spectator ions for the following

yawa3891 [41]

Answer:

Explanation:

1) ZnBr₂ (aq) + AgNO₃ (aq)

Chemical equation:

ZnBr₂ (aq) + AgNO₃ (aq) →Zn(NO₃)₂(aq) + AgBr(s)

Balanced chemical equation:

ZnBr₂ (aq) + 2AgNO₃ (aq) →Zn(NO₃)₂(aq) + 2AgBr(s)

Ionic equation:

Zn²⁺(aq) + Br₂²⁻ (aq) + 2Ag⁺ (aq)+ 2NO⁻₃ (aq) → Zn²⁺(aq) +(NO₃)₂²⁻(aq) + 2AgBr(s)

Net ionic equation:

Br₂²⁻ (aq) + 2Ag⁺ (aq) → 2AgBr(s)

The Zn²⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The AgBr can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

2) Ca(OH)₂ (aq) + Na₂SO₄ (aq)

Chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq) → CaSO₄(s) + NaOH(aq)

Balanced chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq) → CaSO₄(s) + 2NaOH(aq)

Ionic equation:

Ca²⁺(aq) + OH₂²⁻ (aq) + 2Na⁺(aq) + SO₄²⁻ (aq) → CaSO₄(s) + 2Na⁺(aq) + 2OH⁻ (aq)

Net ionic equation:

Ca²⁺(aq) + SO₄²⁻ (aq) → CaSO₄(s)

The OH⁻ ((aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The CaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

3) Al(NO₃)₃ (aq) + Na₃PO₄ (aq)

Chemical equation:

Al(NO₃)₃ (aq) + Na₃PO₄ (aq) → Al(PO₄)(s) + NaNO₃ (aq)

Balanced chemical equation:

Al(NO₃)₃ (aq) + Na₃PO₄ (aq) → Al(PO₄)(s) + 3NaNO₃ (aq)

Ionic equation:

Al³⁺(aq) + 3NO⁻₃ (aq) + 3Na⁺(aq) + PO₄³⁻ (aq) → Al(PO₄)(s) + 3Na⁺(aq) + NO⁻₃ (aq)

Net ionic equation:

Al³⁺(aq) + PO₄³⁻ (aq) → Al(PO₄)(s)

The Na⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The Al(PO₄) can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

4) FeSO₄ (aq) + Ba(OH)₂ (aq)

Chemical equation:

FeSO₄ (aq) + Ba(OH)₂ (aq) → BaSO₄(s) + Fe(OH)₂(aq)

The equation is already balanced.

Ionic equation:

Fe²⁺(aq) + SO₄²⁻ (aq) + Ba²⁺(aq) + 2OH⁻ (aq) → BaSO₄(s) + Fe²⁺(aq) + 2OH⁻(aq)

Net ionic equation:

SO₄²⁻ (aq) + Ba²⁺(aq) → BaSO₄(s)

The Fe²⁺ (aq) and OH⁻ (aq) are spectator ions that's why these are not written in net ionic equation. The BaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

8 0

3 years ago

Give me a hand，Solve this problem，Need everyone's help

elixir [45]

Answer:

A. 8.8 g.

B. 5.3 g.

C. 449 g.

Explanation:

A. Determination of the mass of CO2.

Mole of CO2 = 0.2 mole

Molar mass of CO2 = 12 + (2×16)

= 12 + 32

= 44 g/mol

Mass of CO2 =?

Mole = mass /Molar mass

0.2 = mass of CO2 /44

Cross multiply

Mass of CO2 = 0.2 × 44

Mass of CO2 = 8.8 g

B. Determination of the mass of Na2CO3.

Mole of Na2CO3 = 0.05 mole

Molar mass of Na2CO3 = (2×23) + 12 + (3×16)

= 46 + 12 + 48

= 106 g/mol

Mass of Na2CO3 =?

Mole = mass /Molar mass

0.05 = mass of Na2CO3 /106

Cross multiply

Mass of Na2CO3 = 0.05 × 106

Mass of Na2CO3 = 5.3 g

C. Determination of the mass of Fe(OH)2.

Mole of Fe(OH)2 = 5 moles

Molar mass of Fe(OH)2 = 55.8 + 2(16 + 1)

= 55.8 + 2(17)

= 55.8 + 34

= 89.8 g/mol

Mass of Fe(OH)2 =?

Mole = mass /Molar mass

5 = mass of Fe(OH)2/89.8

Cross multiply

Mass of Fe(OH)2 = 5 × 89.9

Mass of Fe(OH)2 = 449 g.

8 0

3 years ago