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2 years ago

What is the molarity of a solution that contains 2.35 g of nh3 in 0.0500 l of solution?

Chemistry

2 answers:

torisob [31]2 years ago

5 0

Answer:

M = 2.76 M

Explanation:

To calculate the molarity of any solution, we need to use the following expression:

M = n/V (1)

Where:

M: Molarity of solution (mol/L or M)

n: moles of NH3 (moles)

V: volume of solution (In Liters)

But we do not have the moles of NH3, so it needs to be calculated. We have the mass and we can use the following expression to calculate the moles:

n = m/MM (2)

Where:

m: mass of NH3 (g)

MM: molar mass of NH3 (g/mol)

The reported molar mass of NH3 is 17.031 g/mol, so, using the mass of NH3 and this MM, we can calculate the moles:

n = 2.35 / 17.031 = 0.138 moles

With these moles, we can calculate the molarity:

M = 0.138 / 0.05

M = 2.76 mol/L

And this is the concentration or molarity of the NH3 solution.

tankabanditka [31]2 years ago

3 0

The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M

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Butane (c4h10) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[c4h10(g)] = –124.7

vagabundo [1.1K]

The value of Δ H butane (g) = -124.7 kJ/mol

The value of Δ H CO2 (g) = -393.5 kJ/mol

The value of Δ H H2O (g) = -241.8 kJ/mol

Mass of butane, m = 8.30 gm

Molar mass of butane is 58 gm/mol

Consider the reaction,

C₄H₁₀ + 6.5 O₂ = 4CO₂ + 5H₂O

Calculating the value of Δ H° rxn:

ΔH°rxn = ∑nH° f (products) - ∑nH° f (reactants)

Substituting the values we get,

Δ H° rxn = 4 (-393.5) + 5 (-241.8) - (-124.7)

= -1574 -1209 + 124.7

= -2783 - 124.7

= -2658.3 kJ/mol

Now, calculate the number of moles of butane in 8.30 gm.

Number of moles = mass/molar mass

= 8.30 / 58

= 0.143 moles

Thus, the total energy released in the reaction is,

Q = number of moles × ΔH° rxn

= 0.143 × (2658.3)

= 380.14 kJ

Hence, the total heat released in the reaction is 380.14 kJ.

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2 years ago

Mrs. Jacobson is pushing a fridge to the right with a force of 100 N. The force pushing it

alex41 [277]

Answer:

Explanation:

Let's illustrate this; see the attachment.

We see that Mrs. Jacobson is pushing to the right with a force of 100 N and there is another opposite force pushing with a force of 15 N. Since these are in opposite directions, we can say that the force opposite to Mrs. Jacobson is pushing the fridge -15 N to the right (instead of 15 N to the left).

The net force would then be:

100 N + (-15 N) = 85 N to the right

The answer is A.

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2 years ago

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A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

elena55 [62]

Answer:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Explanation:

Mass of an alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Mass of water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

$m_{a}$ $C_{a}$ [ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

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2 years ago

The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is add

JulijaS [17]

Answer:

a)The Ksp was found to be equal to 13.69

Explanation:

Terminology

Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.

Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.

Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).

Step-by-step solution:

To solve this:

#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.

#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.

#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.

a) The equation of solubility equilibrium for KCL is thus;

KCL_(s) ---> K+(aq) + Cl- (aq)

The solubility of KCl given is 3.7 M.

Ksp= [K+][Cl-] = (3.7)(3.7) =13.69

The Ksp was found to be equal to 14.

In pure water KCl

Ksp =13.69 KCl =[K+][Cl-]

Let x= molar solubility [K+],/[Cl-] :. × , x

Ksp =13.69 = [K+][Cl-] = (x)(x) = x²

x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio

37M moles/L

The Ksp was found to be equal to 14.

4.0 M HCl = KCl =[K+][Cl-]

Let y= molar solubility :. y, y+4

Ksp =13.69= [K+][Cl-] = (y)(y*+4)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)

13.69=4y:. y= 3.42 moles/100mL

y= 34.2moles/L

8 M HCl = KCl =[K+][Cl-]

Let b= molar solubility :. B, b+8

Ksp =13.69= [K+][Cl-] = (b)(b*+8)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)

13.69=8b:. b= 1.71 moles/100mL

17.1 moles/L

Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.

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Answer:

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