What can the bromine be classified as

in an experiment 3.5g of element A reacted with 4.0g of element G to form a compound Calculate the empirical formula for this co

kolezko [41]

Additional information

Relative atomic mass(Ar) : A=7, G=16

The empirical formula : A₂G

<h3>Further explanation</h3>

Given

3.5g of element A

4.0g of element G

Required

the empirical formula for this compound

Solution

The empirical formula is the smallest comparison of atoms of compound forming elements.

The empirical formula also shows the simplest mole ratio of the constituent elements of the compound

mol of element A :

$\tt mol=\dfrac{mass}{Ar}\\\\mol=\dfrac{3.5}{7}=0.5$

mol of element G :

$\tt mol=\dfrac{4}{16}=0.25$

mol ratio A : G = 0.5 : 0.25 = 2 : 1

4 0

3 years ago

Hello, I study Home MS Physical Science B-IC. Are there quizlets I can look up for the cumulative exam I have to take today?

Ludmilka [50]

Yes if you search up your subject or topic then put quizlet you’ll find your answer but you may need to login in to get the best experience of studying that you want

8 0

3 years ago

Do all titrations of a strong base with a strong acid have the same ph at the equivalence point?

Luba_88 [7]

Answer:

Yes, all titrations of a strong base with a strong acid have the same pH at the equivalence point.

This pH is 7.

Explanation:

Strong acids and strong bases ionize completely in aqueous solutions. The ionization of strong acids produce hydronium ions, H₃O⁺, and the ionization of strong bases produce hydroxide ions, OH⁻.

Since the ionization of strong acids and bases progress until completion, there is not reverse reaction.

The definition of pH is pH = - log [H₃O⁺]. Acids have low pH (below 7, and greater than 0) and bases have high pH (above 7 and less than 14). Neutral solutions have pH = 7.

Acid-base titrations are a method to determine the concentration of an acid from the known concentration of a base, or the concentraion of a base from the known concentration of an acid.

The equivalence point of the titration is the point at which the the number of moles of hydronium ions and hydroxide ions are equal.

Then, at that point, the hydronium and hydroxide ions will be in the stoichiometric proportion to form a neutral solution, i.e. the pH of the solution wiill be 7.

7 0

3 years ago

What is the name of the compound (NH4)2SO4

GREYUIT [131]

The answer is Ammonium sulfate

3 0

3 years ago

a 25.0-ml volume of a sodium hydroxide solution requires 19.6 ml of a 0.189 m hydrochloric acid for neutralization. a 10.0- ml v

Rashid [163]

Concentration of NaOH = 0.148 molar, M

Concentration of H3PO4 = 0.172 molar, M

Concentration x Volume will give the number of moles of solute in that volume. C*V = moles

Concentration has a unit of (moles/liter). When multiplied by the liters of solution used, the result is the number of moles.

Original HCl solution: (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl

The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl. The reaction is:

NaOH + HCl > NaCl and H2O

This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each. We just determined that the moles of HCl used was 0.00370 moles. Therefore, the 25.0 ml solution of NaOH had the same number of moles: 0.00370 moles NaOH.

The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters). The concentration of NaOH is therefore:

(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M

====

The phosphoric acid problem is handled the same way, but with an added twist. Phosphoric acid is H3PO4. We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4. But now the acid has three hydrogens that will react. The balanced equation for this reaction is:

H3PO4 + 3NaOH = Na3PO4 + 3H2O

Now we need three times the moles of NaOH to neutralize 1 mole of H3PO4.

The moles of NaOH that were used is:

(0.148M)*(0.0349 liters) = 0.00517 moles of NaOH

Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.

The 0.00172 moles of H3PO4 was contained in 10.0 ml. The concentration is therefore:

(0.00172 moles H3PO4)/(0.010 liters H3PO4)

Concentration of H3PO4 = 0.172 molar, M

5 0

1 year ago