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andreev551 [17]
3 years ago
8

What is the [H₃O⁺] (hydronium concentration) of an aqueous solution that has a pH of 4.00?​

Chemistry
1 answer:
Rudik [331]3 years ago
8 0

Answer: 0.0001M

Explanation:

Given that:

[H₃O⁺] (hydronium concentration) = ?

pH of aqueous solution = 4.00

pH is the negative logarithm of hydrogen ion [H+] or hydroxonium ion [H₃O⁺] in a solution. Mathematically it is expressed as:

pH = -log[H₃O⁺]

4.00 = -log[H₃O⁺]

[H₃O⁺] = Antilog (-4.00)

[H₃O⁺] = 0.0001M

Thus, 0.0001M is the [H₃O⁺] (hydronium concentration) of an aqueous solution that has a pH of 4.00

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How many grams of Na2SO4 can be produced from 423.67 g of NaCl?
san4es73 [151]

Mass of  Na2SO4= 514.18 grams

<h3>Further explanation</h3>

Given

423.67 g of NaCl

Required

mass of  Na2SO4

Solution

Reaction

2NaCl + H2SO4 → Na2SO4 + 2HCl

mol NaCl :

= 423.67 g : 58.5 g/mol

= 7.24

From the equation, mol Na2SO4 :

= 1/2 x mol NaCl

= 1/2 x 7.24

= 3.62

Mass  Na2SO4 :

= 3.62 mol x 142,04 g/mol

= 514.18 grams

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3 years ago
2. Which equation represents a transmutation?
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a = Proton transfer d = SN2 Nucleophilic substitution g = Nucleophilic subs at carbonyl(acyl Xfer) b = Lewis acid/base e= Electr
marissa [1.9K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution to this question is shown on the second uploaded image

Explanation:

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3 years ago
photosynthesis in plants require chloroplasts and light energy. identify two materials plants also use in this process ​
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How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of
alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

E = k \frac{q}{r^{2} }

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

E = k \frac{q}{r^{2} }

q = <u>E x r²</u>

        k

q =  <u>1250 N/C x 0.205m</u>²

       8.99 x 10⁹ Nm²/C²

q =   5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

n = \frac{q}{e}

n = <u>5.84 x 10⁻⁹ C </u>

       1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

3 0
3 years ago
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