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.
NaF, BaSO4, KBr AND NaOH are salts
CH4 is an organic compound
NH3 are base
HNO3 is 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
NaOH + H3PO4 --> Na3PO4 NaOH + H3PO4 -->
NaOH + CH3COOH = CH3COONa + H2O Good Luck !
When Na2CO3 reacts with Nitric acid, The products at first are Sodium Hydrogen Carbonate and Sodium Nitrate... HNO3 + Na2CO3 --> NaNO3 + NaHCO3 If again The leftover Sodium hydrogen carbonate is made to react with Nitric Acid, then the products will be: HNO3 + NaHCO3 --> NaNo3 + H2O + CO2
NaOH + CO2--------------- NaHCO3 sodium bicarbonate aka Baking Soda
NaOH + HNO3 -----> NaNO3 + H2O
NaOH + HNO3 -> NaNO3 + H2O
HNO3+ NaOH = NaNO3+ H2O is a neutralization reaction
No. BaSO4 is a salt. NaOH is a base.
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
Yes. The reaction is fast and exothermic. The equation is NaOH + HNO3 --> H2O + NaNO3
NaNO3 and water
Yes.
0.289 Moles is the molarity of an NaOH solution if 4.37 ml is titrated by 11.1 ml of 0.0904m hno3.
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 )
HNO3 + NaOH NaNO3 + H2O
NaNO3 is a salt which is produced by the reaction of HNO3 and NaOH