When 35.47 g of sodium hydroxide react with boric acid (H3BO3), how many moles of sodium borate will be produced?

5.83 g

Explanation:

First, you must start with a balanced equation so you can see the mole ratios.

NaOH + H₃BO₃ --> NaBO₂ + 2H₂O

You can see that it takes 1 mole of sodium hydroxide to form 1 mole of sodium borate. 1:1 ratio

Now you must calculate how many moles of NaOH 35.47 g equals.

Na = 22.99 amu

O = 15.99 amu

H = 1.008 amu

NaOH = 39.997 amu

35.47 g ÷ 39.997 amu = 0.08868 moles of NaOH

Since it's a 1:1 ratio, the same number of moles of NaBO₂ is created. Now you must convert moles to grams.

Na = 22.9 amu

B = 10.81 amu

2 O = 31.998 amu

NaBO₂ = 65.798 amu

0.08868 moles x 65.798 = 5.83 g

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

o conteúdo de calor de um sistema é igual à entalpia apenas para um sistema que é constante