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CH4 an organic alkane. (Methane)
NaF An inorganic salt ( Sodium fluoride)
NaOH A soluble base / alkali ( Sodium hydroxide)
BaSO4 An inorganic salt ) Barium sulphate)
NH3 A basic gas ( Ammonia)
KBr An inorganic salt (Potassium bromide)
HNO3 A mineral acid (Nitric acid)
NB
When writing chemical formula remember single letter elements use CAPITAL letter. Two-letter elements have first letter as a capital and the second letter as small/lower case. This is standard world wide practice in order to avoid confusion over names.
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How many moles of sodium hydroxide are required to neutralize 0.20 mol of nitric acid?
Balanced Equation. NaOH + HNO3 >> NaNO3 + H2O Now, Molarity = moles of solute/liters of solution 0.800M HNO3 + mol/2.50L mol of HNO3 = 2 2mol HNO3 (1mol NaOH/1molHNO3 )(39.998g NaOH/1mol NaOH ) = 79.996 grams
What is the formula of two salts formed from sodium hydroxide and phosphoric acid?
NaOH + H3PO4 --> Na3PO4 NaOH + H3PO4 -->
What is balanced chemical equation for reaction between naoh plus ch3cooh?
NaOH + CH3COOH = CH3COONa + H2O Good Luck !
What is the reaction equation between NH3 and HNO3?
In this case, this is an acid-base reaction between nitric acid and ammonia. Nitric Acid is a strong acid, therefore, its hydrogen atom dissociates completely. The equation looks like this:HNO3(aq) + NH3(aq) => NH4NO3(aq)
What is the name of the polyatomic ion NaOH?
NaOH is Sodium Hydroxide.
HNO3 + NaOH NaNO3 + H2O?
HNO3+ NaOH = NaNO3+ H2O is a neutralization reaction
What is the acid base reaction for HNO3 plus NaOH?
The acid-base reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) produces sodium nitrate (NaNO3) and water (H2O). The balanced chemical equation for this reaction is: HNO3 + NaOH → NaNO3 + H2O.
What is the molarity of HNO3 if 20.0 ml of the solution is needed to exactly neutralize 10.0 ml of a 1.67 M NaOH solution?
The reaction between HNO3 and NaOH is a 1:1 molar ratio. This means that the moles of HNO3 required to neutralize the NaOH is the same as the moles of NaOH. Given that 20.0 ml of HNO3 is needed to neutralize 10.0 ml of a 1.67 M NaOH solution, the molarity of the HNO3 solution is twice the molarity of the NaOH solution, which is 3.34 M.
What is the molarity of a nitric acid solution HNO3 if 20.0 mL of the solution is needed to exactly neutralize 10.0 mL of a 1.67 M NaOH solution?
The balanced chemical equation for the neutralization reaction is: HNO3 + NaOH → NaNO3 + H2O From the equation, we know that the mole ratio of HNO3 to NaOH is 1:1. This means that the moles of HNO3 will be equal to the moles of NaOH. First, calculate the moles of NaOH used: Moles NaOH = (10.0 mL) x (0.001 L/mL) x (1.67 mol/L) = 0.0167 mol Since the moles of NaOH are equal to the moles of HNO3, we have 0.0167 mol of HNO3 in 20.0 mL of solution: Molarity of HNO3 = moles HNO3 / volume of solution (L) = 0.0167 mol / 0.020 L = 0.835 M
What is the molarity of an naoh solution if 4.37ml is titrated by 11.1 ml of 0.0904 m hno3?
The balanced equation for the reaction is 1 mole of NaOH to 1 mole of HNO3. Using the titration data, you can calculate the moles of HNO3 used. From there, you can determine the moles of NaOH present in the 4.37 ml solution. Finally, dividing the moles of NaOH by the volume of the NaOH solution in liters will give you the molarity.
How many moles of sodium hydroxide are required to neutralize 0.20 mol of nitric acid?
Balanced Equation. NaOH + HNO3 >> NaNO3 + H2O Now, Molarity = moles of solute/liters of solution 0.800M HNO3 + mol/2.50L mol of HNO3 = 2 2mol HNO3 (1mol NaOH/1molHNO3 )(39.998g NaOH/1mol NaOH ) = 79.996 grams
What is the equation of a reaction of nitric acid and sodium hydroxide?
HNO3 (aq) + NaOH (aq) --> H2O (l) + NaNO3 (aq)
What is the balanced equation for the reaction of nitric acid and sodium hydroxide?
Nitric acid: HNO3 (acid) Sodium hydroxide: NaOH (base) This is therefore an acid-base reaction. Acid + Base --> Salt + Water Therefore: HNO3 + NaOH --> NaNO3 + H20 Or: Nitric acid + Sodium hydroxide --> Sodium Nitrate + Water
A nitric acid soulution is neutralized using sodium hydroxide how many grams of sodium hydroxide are needed to neutralize 2.50 L of 0.800 M Nitriric acid soulution?
Balanced equation first. HNO3 + NaOH >> NaNO3 + H2O Now, Molarity = moles of solute/volume of solution ( find moles HNO3 ) 0.800 M HNO3 = moles/2.50 Liters = 2 moles of HNO3 ( these reactants are one to one, so we can proceed to grams NaOH ) 2 moles NaOH (39.998 grams NaOH/1mole NaOH) = 79.996 grams of NaOH need to neutralize the acid. ( you do significant figures )
What will happen if NaOH plus HNO3?
When NaOH (sodium hydroxide) reacts with HNO3 (nitric acid), a neutralization reaction occurs to form water and sodium nitrate (NaNO3). This reaction releases heat and is exothermic due to the strong acidic and basic nature of the reactants. The products formed are salt (NaNO3) and water.
Complete ionic equation for HNO3-NaOH?
the answer for the reaction HNO3 AND NaOH is :- NaNo3 + H20 that coz when acid reacts with a metal oxide it produces salt and water --> all metals are alkali so if acid reacts with either metaloxide or alkali it produces salt and water... hope that helped.. ;)
What salt is produced by mixing NaOH and HNO3?
When sodium hydroxide (NaOH) reacts with nitric acid (HNO3), the salt produced is sodium nitrate (NaNO3). Additionally, water is also formed as a byproduct in this neutralization reaction.
