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qaws [65]
2 years ago
7

Now you will solve the same problem as above, but using the quadratic formula instead of iterations, to show that the same value

of x is obtained either way. using the quadratic equation to calculate [h3o+] in 0.00250 m hno2, what are the values of a, b, c and x , where a, b, and c are the coefficients in the quadratic equation ax2+bx+c=0, and x is [h3o+]? recall that ka=4.5×10−4 .
Chemistry
1 answer:
Dafna1 [17]2 years ago
6 0

Answer:

a: 1

b: 4.5x10⁻⁴

c: 1.125x10⁻⁶

[H₃O⁺] = 0.000859M

Explanation:

As HNO₂ is a weak acid, its equilibrium in water is:

HNO₂(aq) + H₂O(l) ⇄ H₃O⁺(aq) + NO₂⁻(aq)

Equilibrium constant, ka, is defined as:

ka = 4.5x10⁻⁴ = [H₃O⁺] [NO₂⁻] / [HNO₂] <em>(1)</em>

Equilibrium concentration of each specie are:

[HNO₂] = 0.00250M - x

[H₃O⁺] = x

[NO₂⁻] = x

Replacing in (1):

4.5x10⁻⁴ = x × x / 0.00250M - x

1.125x10⁻⁶ - 4.5x10⁻⁴x = x²

0 = x² + 4.5x10⁻⁴x - 1.125x10⁻⁶

As the quadratic equation is ax² + bx + c = 0

Coefficients are:

a: 1

b: 4.5x10⁻⁴

c: 1.125x10⁻⁶

Now, solving quadratic equation:

x = -0.0013 → False answer, there is no negative concentrations.

<em>x = 0.000859</em>

As [H₃O⁺] = x; <em>[H₃O⁺] = 0.000859M</em>

I hope it helps!

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Which of the following is a correctly written chemical equation that demonstrates the conservation of mass?
Fantom [35]

Answer:

Option D is correct.

H₂O + CO₂      →    H₂CO₃

Explanation:

First of all we will get to know what law of conservation of mass states.

According to this law, mass can neither be created nor destroyed in a chemical equation.

This law was given by French chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Example:

6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂

there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.

Now we will apply this law to given chemical equations:

A) H₂ + O₂   →    H₂O

There are two hydrogen and two oxygen atoms present on left side while on right side only one oxygen and two hydrogen atoms are present so mass in not conserved. This equation not follow the law of conservation of mass.

B) Mg + HCl   →   H₂ + MgCl₂

In this equation one Mg, one H and one Cl atoms are present on left side while on right side two hydrogen, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.

C) KClO₃      →     KCl + O₂

There are one K, one Cl and three O atoms are present on left side of chemical equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.

D)  H₂O + CO₂      →    H₂CO₃

There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. Thus is correct option.

6 0
2 years ago
Before landing, the brakes and the tires of an airliner have a temperature of 15.0∘C. Upon landing, the 90.7 kg carbon fiber bra
Goryan [66]

Answer:

0.921 J/g degrees C

Explanation:

Recall that the First Law of Thermodynamics demands that the total internal energy of an isolated system must remain constant. Any amount of energy lost by the brakes must be gained by the tires (in the form of heat in this situation).  Therefore, heat given off by the brakes = −heat taken in by tires, or:

−qbrakes=qtires

The equation used to calculate the quantity of heat energy exchanged in this process is:

−qbrakes=−cbrakes mbrakes ΔTbrakes=ctires mtires ΔTtires=qtires

First we must convert the mass of the tires and the brakes from  kg to  g.

massbrakes=90.7 kg×1,000. g1 kg=9.07×104 g

masstires=123 kg×1,000. g1 kg=1.23×105 g

Next, substitute in known values and rearrange to solve for ctires. Note that the final temperature for both the tires and the brakes is 172∘C, the initial temperature of the brakes is 312∘C and the initial temperature of the tires is 15∘C.

−(1.400Jg∘C)(9.07×104 g)(172∘C−312∘C)=(ctires)(1.23×105 g)(172∘C−15∘C)

ctires=−(1.400 Jg∘C)(9.07×104 g)(−140∘C)(1.23×105 g)(157∘C)=17,777,200 J19311000 g∘C=0.9206Jg∘C

The answer should have three significant figures, so round to 0.921Jg∘C.

6 0
2 years ago
Which of the following compounds would be most effective in lowering the melting point of ice on roads? a) CaCl2 b) NaCl C) K3PO
Mrac [35]
I'm not 100% sure on this, but I would go with C) NaCl. 
NaCl is a salt, and that is used to melt the ice on the roads. Hope this helps!
7 0
2 years ago
If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the
jeyben [28]

25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

<u>Explanation:</u>

2 NH₃ + H₂S ----> (NH₄)₂S​

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S​ = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S​

1 g of NH₃ will produce \frac{68}{34} g of (NH₄)₂S​

25g of NH₃ will produce \frac{65}{34} X 25 g of (NH₄)₂S​

                                     = 47.8 g of (NH₄)₂S​

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

4 0
2 years ago
The specific heat of aluminum is 0.214 cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5
Natasha2012 [34]
For this problem, we use the formula for sensible heat which is written below:

Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference

Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
4 0
2 years ago
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