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3 years ago

67.4 L of sulfur dioxide (SO2) gas at 0.0C and at 1 atm of pressure is equivalent aren’t to how many moles?

Chemistry

1 answer:

gizmo_the_mogwai [7]3 years ago

6 0

Solution here,

Volume(V)=67.4 L

Pressure(P)=1 atm

Temperature(T)=(0+273)K=273K

Universal gas constant(R)=0.0821 L.atm.mol^-1K^-1

No. of moles(n)=?

Now,

PV=nRT

or, 1×67.4=n×0.0821×273

or, 67.4=22.4n

or, n=67.4/22.4

or, n=3

therefore, required no. of mole is 3.

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What happens when sodium trioxocarbonate (IV) decahydrate is heated

PSYCHO15rus [73]

Answer:

the simplest answer is it loses the water (decahydrate) because it evaporates

5 0

2 years ago

Write the complete ionic equations, spectator ions and net ionic equation for the following.

gizmo_the_mogwai [7]

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) HgCl₂ (aq) + KI (aq) →

Chemical equation:

HgCl₂ (aq) + KI (aq) → KCl + HgI₂

Balanced chemical equation:

HgCl₂ (aq) + 2KI (aq) → 2KCl(aq) + HgI₂(s)

Ionic equation:

Hg²⁺(aq) + Cl₂²⁻ (aq) + 2K⁺(aq) + 2I⁻ (aq) → HgI₂ (s) + 2K⁺(aq) + 2Cl⁻ (aq)

Net ionic equation:

Hg²⁺(aq) + 2I⁻ (aq) → HgI₂ (s)

The Cl⁻ ((aq) and K⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The HgI₂ (s) 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) 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.

4 0

3 years ago

How do I do question number 3? And what is the answer?

shutvik [7]

Answer:

60.02 g.

Explanation:

It is clear from the balanced equation:

Mg + 2HCl → MgCl₂ + H₂.

that 1.0 mole of Mg reacts with 2.0 moles of HCl to produce 1.0 mole of MgCl₂ and 1.0 moles of H₂.

20.0 g of Mg reacts with excess HCl. To calculate the no. of grams of HCl that reacted, we should calculate the no. of moles of Mg:

no. of moles of Mg = mass/atomic mass = (20.0 g)/(24.3 g/mol) = 0.823 mol.

From the balanced equation; every 1.0 mol of Mg reacts with 2 moles of HCl.

∴ 0.833 mol of Mg will react with (2 x 0.833 mol = 1.646 mol) of HCl.

∴ The mass of reacted HCl = no. of moles x molar mass = (1.646 mol)(36.46 g/mol) = 60.02 g.

3 0

3 years ago

Select the correct hybridization for the central atom based on the electron geometry cocl2 (carbon is the central atom).

Rainbow [258]

In this compound (Phosgene) the central atom (carbon is Sp² Hybridized).

Sp, Sp² and Sp³ can be calculated very simply by doing three steps,

Step 1:
Assume triple bond and double bond as one bond and assign s or p to it. In this example carbon double bond oxygen is considered once and let suppose it is s. Now we are having our s.

Step 2:
Count lone pair of electron, each lone pair counts for s and p. In this case there is no lone pair of electron on carbon, so not included.

Step 3:
Count single bonds for s and p. As we have already assigned s to the double bond, now one p for one single bond, and other p for the other single bond.

Result:
So, we counted 1 s for double bond, 1 p for one single and other p for second single bond. As a whole we got,

Sp²

Practice:
You can practice for hybridization of Oxygen in this molecule. Oxygen has 2 lone pair of electrons. (Hint: Sp² Hybridization)

7 0

3 years ago

Tissue that makes up the outer covering in humans

Rainbow [258]

I think it's called skin

4 0

3 years ago

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