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3 years ago

Determine the ph of a 0.500 m hno2 solution. Ka of hno2 is 4.6 × 10−4.

Chemistry

1 answer:

Mumz [18]3 years ago

3 0

Answer: pH of a 0.500 M $HNO_2$ solution is 1.82

Explanation:

$HNO_2\rightarrow H^++NO_2^-$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.500 M and $\alpha$ = ?

$K_a=4.6\times 10^{-4}$

Putting in the values we get:

$4.6\times 10^{-4}=\frac{(0.500\times \alpha)^2}{(0.500-0.500\times \alpha)}$

$(\alpha)=0.030$

$[H^+]=c\times \alpha$

$[H^+]=0.500\times 0.030=0.015$

Also $pH=-log[H^+]$

$pH=-log[0.015]=1.82$

Thus pH of a 0.500 M $HNO_2$ solution is 1.82

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Is the genie in the bottle experiment a physical or chemical change/reaction?

gladu [14]

Answer:

Chemical reaction.

Step-by-step explanation:

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2H₂O₂(ℓ) ⟶ 2H₂O(ℓ) + O₂(g)

New substances are formed, and old ones disappear, so this is a chemical reaction.

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The rapidly-expanding oxygen gas caries with it droplets of water and hydrogen peroxide (and probably some steam, as well).

The visual effect is like a genie escaping from its bottle.

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3 years ago

Which of the following terms means that metals can be pulled into thin strands or wires? [like copper]

Dimas [21]

Answer:

Explanation:

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3 years ago

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A substance decays so that the amount a of the substance left after t years is given by: a = a 0 · (0.9) t , where a 0 is the or

Dennis_Churaev [7]

I think the correct form of the equation is given as:

a = a0 * (0.9)^t

where t is an exponent of 0.9 since this is an exponential decay of 1st order reaction

Now to solve for the half life, this is the time t in which the amount left is half of the original amount, therefore that is when:

a = 0.5 a0

Substituting this into the equation:

0.5 a0 = a0 * (0.9)^t

0.5 = (0.9)^t

Taking the log of both sides:

t log 0.9 = log 0.5

t = log 0.5 / log 0.9

t = 6.58 years

Answer:

half life = 6.58 years

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3 years ago

(1.) Using Beer's Law, How will the absorbance measured for the solutions change as the concentration of aspirin in solutions in

Vesnalui [34]

Answer:

(1) The absorbance of the aspirin in solutions will increase.

(2) [ASA]f = 3.79x10⁻⁴M

(3) [ASA]i = 3.79x10⁻³M

(4) m ASA = 0.171g

Explanation:

The Beer's Law is expressed by:

$A = \epsilon \cdot l \cdot C$ (1)

where A: is the absorbance of the species, ε: is the molar attenuation coefficient, l: is the pathlength and C: is the concentration of the species

(1) From equation (1), the relation between the absorbance of the species and its concentration is directly proportional, so if the aspirin concentration in solutions increases, the absorbance of the solutions will also increase.

(2) Starting in the given expression for the relationship between absorbance and concentration of ASA, we can calculate its concentration in the solution:

$A = 1061.5 \cdot [ASA]$

$[ASA] = \frac{A}{1061.5} = 3.79 \cdot 10^{-4}M$

Therefore, the aspirin concentration in the solution is 3.79x10⁻⁴ M

(3) To calculate the stock solution concentration, we can use the next equation:

$V_{i} [ASA]_{i} = V_{f} [ASA]_{f}$

where Vi: is the stock solution volume=10mL, Vf: is the solution diluted volume=100mL, [ASA]i: is the aspirin concentration of the stock solution and [ASA]f: is the aspirin concentration of the diluted solution

$[ASA]_{i} = \frac{V_{f} \cdot [ASA]_{f}}{V_{i}} = \frac {100mL \cdot 3.79\cdot 10^{-4} M}{10mL} = 3.79 \cdot 10^{-3} M$

Hence, the concentration of the stock solution is 3.79x10⁻³M

(4) To determine the aspirin mass in the tablet, we need to use the following equation:

$m_{ASA} = \eta_{ASA} \cdot M_{ASA} = [ASA]_{i} \cdot V_{0} \cdot M_{ASA}$

where η: is the aspirin moles = [ASA]i V₀, M: is the molar mass of aspirin=180.158g/mol, V₀: is the volume of the volumetric flask=250mL and [ASA]i: is the aspirin concentration in the volumetric flask which is equal to the stock solution=3.79x10⁻³M

$m_{ASA} = 3.79 \cdot 10^{-3} \frac{mol}{L} \cdot 0.250L \cdot 180.158 \frac{g}{mol} = 0.171 g$

Then, the aspirin mass in the tablet is 0.171 g.

I hope it helps you!

5 0

3 years ago

If a 1.271-g sample of aluminum metal is heated in a chlorine gas atmosphere,

Gala2k [10]

Answer:

AlCl₃

Explanation:

Data Given:

Mass of aluminum metal = 1.271 g

Mass of aluminum chloride = 6.280 g

Empirical formula of aluminum chloride = ?

Solution:

First find mass of Chlorine

As 6.280 g of aluminum chloride produced by 1.271 g so

the mass of chlorine in 6.280 g will be 6.280 g -1.271 g)

Mass of chlorine = 5.009 g

Now

Find the number of moles of Al and chlorine in aluminum chloride.

Molar Mass of Al = 26.98 g/mole

Molar mass of Cl = 35.5 g/mol

Mole of Al

Formula Used

no. of moles = mass in grams / molar mass . . . . . . . . (1)

Put values in equation 1

no. of moles of Al = 1.271 g / 26.98 g/ mole

no. of moles of Al = 0.0471

Mole of Chlorine

Formula used

no. of moles = mass in grams / molar mass . . . . . . . . (1)

Put values in equation 1

no. of moles of Cl = 5.009 g / 35.5 g/ mole

no. of moles of Cl = 0.1411

Now we have

Al = 0.0471 moles

Cl = 0.1411 moles

As we Know

Empirical formula shows the simplest ratio of atoms in the molecule but not whole numbers of atoms in a compound.

So,

The ratio of moles of Al to chlorine is

Al : Cl

0.0471 0.1411

Divide the ratio by smallest number to get simplest whole number ratio

Al : Cl

0.0471 / 0.0471 0.1411/ 0.0471

Al : Cl

1 : 3

The simplest ratio of Al to cl is 1 to 3, so the formula will be

Emperical formula of aluminum chloride = AlCl₃

7 0

3 years ago