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Anton [14]
2 years ago
10

What volume of a 0.00945-M solution of potassium hydroxide would be required to titrate 50.00 mL of a sample of acid rain with a

H2SO4 concentration of 1.23 × 10−4 M. H2SO4(aq)+2KOH(aq)⟶K2SO4(aq)+2H2O(l)
Chemistry
1 answer:
andrezito [222]2 years ago
3 0

Answer:

1.30mL

Explanation:

The equation for the reaction is given below:

H2SO4 + 2KOH —> K2SO4 + 2H2O

From the equation above we obtained the following:

Mole of acid (nA) = 1

Mole of base (nB) = 2

The following data were obtained from the question:

Mb = 0.00945M

Vb =?

Va = 50mL

Ma = 1.23 × 10^−4M

Using MaVa / MbVb = nA/nB, we can calculate the volume of KOH as illustrated below:

MaVa / MbVb = nA/nB

(1.23 × 10^−4 x 50)/0.00945xVb = 1/2

Cross multiply to express in linear form

1.23 × 10^−4 x 50 x 2 = 0.00945xVb

Divide both side by 0.00945

Vb = (1.23 × 10^−4 x 50 x 2) /0.00945

Vb = 1.30mL

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Some hydrogen gas is enclosed within a chamber being held at 200∘C∘C with a volume of 0.0250 m3m3. The chamber is fitted with a
Semenov [28]

Answer:

The final volume is 39.5 L = 0.0395 m³

Explanation:

Step 1: Data given

Initial temperature = 200 °C = 473 K

Volume = 0.0250 m³ = 25 L

Pressure = 1.50 *10^6 Pa

The pressure reduce to 0.950 *10^6 Pa

The temperature stays constant at 200 °C

Step 2: Calculate the volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.50 * 10^6 Pa

⇒with V1 = the initial volume = 25 L

⇒with P2 = the final pressure = 0.950 * 10^6 Pa

⇒with V2 = the final volume = TO BE DETERMINED

1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2

V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)

V2 = 39.5 L = 0.0395 m³

The final volume is 39.5 L = 0.0395 m³

3 0
2 years ago
In 200 g of a concentrated solution of 70.4 wt% nitric acid (r = 1.41 g/mL, FW(HNO3) = 63.01 g/mol), how many grams of water are
mars1129 [50]

Answer:

59.2 grams

Explanation:

We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:

W = 200*(1-0.704)\\W=59.2\ g

There are 59.2 grams of water in this solution.

5 0
2 years ago
When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 6
Mariulka [41]

Answer:

The energy difference between these 2p and 2s orbitals is 3.07\times 10^{-19} J

Explanation:

Wavelength of the photon emitted = \lambda =646.3 nm =646.3\times 10^{-9} m

Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E

Energy of the photon is given by Planck's equation:

E=\frac{hc}{\lambda }

h = Planck's constant = 6.626\tiomes 10^{-34} Js

c = Speed of the light = 3\times 10^8 m/s

E=\frac{6.626\tiomes 10^{-34} Js\times 3\times 10^8 m/s}{646.3\times 10^{-9} m}

E=3.07\times 10^{-19} J

The energy difference between these 2p and 2s orbitals is 3.07\times 10^{-19} J

3 0
2 years ago
If 3.491 grams of the precipitate was formed, how many moles of strontium bromide were reacted
AVprozaik [17]
The balanced chemical equation that represents the reaction is as follows:
 <span>SrBr2(aq) + 2AgNO3(aq) → Sr(NO3)2(aq) + 2AgBr(s) 
</span>
From the periodic table:
mass of silver = 108 grams
mass of bromine = 80 grams

molar mass of silver bromide = 108 + 80 = 188 grams

number of moles = mass / molar mass
number of moles of produced precipitate = 3.491/188 = 0.018 moles

From the balanced equation:
1 mole of  strontium bromide produces 2 moles of silver bromide. Therefore, to calculate the number of moles of <span>strontium bromide that produces 0.018 moles of silver bromide, you will just do a cross multiplication as follows:
amount of </span><span>strontium bromide = (0.018x1) / 2 = 9.28 x 10^-3 moles</span>
4 0
2 years ago
The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super
Katen [24]

Answer:

The pH of the solution is 8.

Explanation:

To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:

1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:

pOH = –Log [OH-]

pOH = 6

6 = –Log [OH-]

–6 = Log [OH-]

[OH-] = Antilog (–6)

[OH-] = 1x10^–6 mol/L

2. The pH of the solution can be obtained as follow:

pH + pOH = 14

pOH = 6

pH + 6 = 14

pH = 14 – 6

pH = 8.

From the calculations made above,

[OH-] = 1x10^–6 mol/L

pH = 8.

Therefore, the correct answer is:

The pH of the solution is 8

3 0
2 years ago
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