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loris [4]
3 years ago
11

Gold ions occur in seawater to the extent of1.0.

Chemistry
1 answer:
faltersainse [42]3 years ago
5 0

Answer:

According to research, the concentration of gold ions in the seawater is 10⁻¹⁴g/cm³.

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Balance the following chemical reaction: _NaOH + _H2SO4 ----->_Na2SO4 +_H20
Dima020 [189]

Answer:

Sodium Hydroxide + Sulfuric Acid = Sodium Sulfate + Water

2NaOH + H2SO4 → Na2SO4 + 2H2O

Explanation:

To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. The balanced equation will appear above.

Use uppercase for the first character in the element and lowercase for the second character. Examples: Fe, Au, Co, Br, C, O, N, F.

Ionic charges are not yet supported and will be ignored.

Replace immutable groups in compounds to avoid ambiguity. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will.

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8 0
3 years ago
PLZ HELP
riadik2000 [5.3K]
Too many to know in the world.
8 0
3 years ago
Calculate the pH during the titration of 30.00 mL of 0.1000 M HCOOH(aq) with 0.1000 M NaOH(aq) after 29.3 mL of the base have be
pashok25 [27]

Answer:

3.336.

Explanation:

<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>

<em />

So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>

<em></em>

<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>

∵ pH = - log[H⁺].

<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>

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What factors affect the temperature of sea water
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The two main factors the temperature of seawater are density and the salinity of the water.
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Is water wet? I give you a lot of points
UkoKoshka [18]

i mean technically, no. only because water is water and water makes things wet. you know? unless you pour water onto water then idk honestly, truly...

6 0
3 years ago
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