95.0 - 17.524 sig fig

2H2 + O2 --> 2H2O can be classified as what type of reaction?

Sveta_85 [38]

Combustion reaction
Key: O2
O2 is normally in a chemical formula when you are used to burn anything, so basically, anything with O2 involves burning.

5 0

3 years ago

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Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na 2CrO 4 are added to 250 mL of 0.12 M A

olchik [2.2K]

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:

Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)

Ksp is defined as:

Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are in equilibrium

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are the actual concentrations

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M

And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

[Ag⁺] = 0.06M - 2X

[CrO₄²⁻] = 0.165M - X

Where X is defined as the reaction coordinate

Replacing in Ksp expression:

1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]

Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

X = 0.0299986M

Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

[CrO₄²⁻] = 0.165M - 0.0299986M

[CrO₄²⁻] = 0.135M

6 0

4 years ago

What is 2-2? When the cow is green

bija089 [108]

0 with a green cow lol

6 0

3 years ago

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What is the sum of the coefficients of the balanced equation for:

Marianna [84]

Answer:

3 Pb(NO3)2 + Al2(SO4)3 = 2 Al(NO3)3 + 3 PbSO4

Explanation:

I think your equation is incorrect? This is balanced and the sum of coefficients is 9.

5 0

3 years ago

The vapor above a mixture of pentane and hexane at room temperature contains 35.5% pentane by mass. What is the mass percent com

FrozenT [24]

The mass percent of Pentane in solution is 16.49%

The mass percent of Hexane in solution is 83.51%

Explanation:

Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.

Convert these values to mol% using their molecular weights:

Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol

Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol

Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%

Hexane mol%: yh = 100 - 39.68 = 60.32%

Pp-vap = 425 torr = 0.555atm

Ph-vap = 151 torr = 0.199atm

From Raoult's law we know:

Pp = xp $\times$ Pp - vap = yp $\times$ Pt (1)

Ph = xh $\times$ Ph - vap = yh $\times$ Pt (2)

Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:

(1 - xp) $\times$ Ph - vap = (1 - yp) $\times$ Pt (3)

Substituting (1) into (3) we get:

(1-xp) $\times$ Ph - vap = (1 - yp) $\times$ xp $\times$ Pp - vap / yp (4)

Rearrange for xp:

xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap) (5)

Subbing in the values we find:

Pentane mol% in solution: xp = 19.08%

Hexane mol% in solution: xh = 80.92%

Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:

mp = 0.1908 mol $\times$ 72.15 g/mol

= 13.766 g

mh = 0.8092 mol $\times$ 86.18 g/mol

= 69.737 g

Mass% of Pentane solution = 13.766/(13.766+69.737)

= 16.49%

Mass% of Hexane solution = 83.51%

8 0

4 years ago