How many oxygen atoms are present in 8.186*10-20 g of hclo4?

Calculate the number of grams in sodium in 7.6g of Na2C6H6O7 (sodium hydrogen citrate) express your answer using two significant

Katen [24]

Answer:

Explanation:

the molecular mass of Na2C6H6O7 =236 g\mole it has a sodium that has 23 g/mole so 7.6 g of Na2C6H6O7 has x of sodium mass

236 g/mole ⇒ 23g/mole

7.6 x 23 ÷ 236 = 74.07×10-2 grams of sodium

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7 0

3 years ago

Write the complete balanced equation for the neutralization reaction that occurs when aqueous hydroiodic acid, HI, and sodium hy

Serhud [2]

Answer:

H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g )

Explanation:

The complete reaction when hydroiodic acid and sodium hydrogen carbonate combine, would be as follows -

HI + NaHCO3 ----> NaI + H2O + CO2

net reaction

H2CO3 is highly unstable, and thus decomposes into the water and carbon dioxide you see present as the reactants. If you didn't know already, H2CO3 is also reffered to as carbonic acid. The rest of the elements present on the reactant side are Iodine and Sodium, which is why they are present on the product side as NaI.

Let me include the " physical states " in this reaction as well -

HI ( aq ) + NaHCO3 ( aq ) ----> NaI ( aq ) + H2O ( l ) + CO2 ( g )

Now the complete ionic equation would simply be each compound present as ions in an aqueous solution, so there is no need for an explanation on this step -

H+ ( aq ) + I- ( aq ) + Na+ ( aq ) + HCO3- ( aq ) -------> Na+ ( aq ) + I- ( aq ) + H2O( l ) + CO2 ( g )

The spectator ions in this reaction are I- and Na+, so canceling them out, you would receive the following net ionic equation -

H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g )

<u><em>Hope that helps!</em></u>

4 0

3 years ago

How many molecules are in 2.00 moles of h2o

Sedaia [141]

1 mole = 6.022×10^23 atoms. 1 water molecule = 2 Hydrogen atoms + 1 oxygen atom. So, 1 mole H2O = 1.2044×10^24 hydrogen atoms. Therefore 2 mole H2O will have 2.4088×10^24 hydrogen atoms.

8 0

3 years ago

Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.

STALIN [3.7K]

Answer: Solution A : $[H_3O^+]=0.300\times 10^{-7}M$

Solution B : $[OH^-]=0.107\times 10^{-5}M$

Solution C : $[OH^-]=0.177\times 10^{-10}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

$pH=-\log[H_3O^+]$