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algol [13]
3 years ago
11

The net ionic equation for the reaction between aqueous nitric acid and aqueous sodium hydroxide is ________. h+ (aq) + na+ (aq)

+ oh- (aq) → h2o (l) + na+ (aq) h+ (aq) + hno3 (aq) + 2oh- (aq) → 2h2o (l) + no3- (aq) h+ (aq) + oh- (aq) → h2o (l) hno3 (aq) + naoh (aq) → nano3 (aq) + h2o (l) hno3 (aq) + oh- (aq) → no3- (aq) + h2o (l)
Chemistry
1 answer:
Nana76 [90]3 years ago
7 0
HNO3+NaOH ----> H2O
H⁺ +NO3⁻+Na⁺+OH⁻ ---> Na⁺ +NO3⁻ +H2O
H⁺ (aq)+OH⁻(aq)----> H2O(l)
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Questions and answers about isotopes grade 8 I need this for a PowerPoint exam projects
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3 0
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How do elements on the right side of the periodic table differ from elements on the left side of the table?
lozanna [386]

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4 0
3 years ago
Give the chemical symbols for the following elements: (a) potassium, (b) tin, (c) chromium, (d) boron,(e) barium, (f) plutonium,
Ilia_Sergeevich [38]

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8 0
4 years ago
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5 0
3 years ago
A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19g/mL at 25°C what is the molarity and molality of the soluti
lilavasa [31]

<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg

<em>Molar concentration </em>

Assume you have 1 L solution.

Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)

= 1190 g solution

Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)

= 84.01 g NaHCO3

Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)

= 1.14 mol NaHCO3

<em>c</em> = 1.14 mol/1 L = 1.14 mol/L

<em>Molal concentration</em>

Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg

<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg

5 0
3 years ago
Read 2 more answers
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