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2 years ago

How many grams do 6.534e+24 molecules of phosphoric acid weigh?

Chemistry

1 answer:

fredd [130]2 years ago

3 0

Answer:

1,063 grams H₃PO₄

Explanation:

To find the mass of phosphoric acid (H₃PO₄), you should (1) convert molecules to moles (via Avogadro's number) and then (2) convert moles to grams (via molar mass from periodic table).

Molar Mass (H₃PO₄): 3(1.008 g/mol) + 30.974 g/mol + 4(15.998 g/mol)

Molar Mas (H₃PO₄): 97.99 g/mol

6.534 x 10²⁴ molecules H₃PO₄ 1 mole 97.99 g
--------------------------------------------- x ------------------------------------- x --------------
6.022 x 10²³ molecules 1 mole

= 1,063 grams H₃PO₄

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A 26.08 g mixture of zinc and sodium is reacted with a stoichiometric amount of sulfuric acid. The reaction mixture is then reac

Over [174]

Answer:

Molar percent of sodium in original mixture is 88,50%

Explanation:

The last reaction is:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl

The moles of BaCl₂ are:

0,132L × 3,80M = 0,502 moles of BaCl₂

As the amount of BaCl₂ is the maximum possible to produce BaSO₄, the moles of BaCl₂ must be the same than moles of Na₂SO₄.

0,502 moles of BaCl₂ ≡ 0,502 moles of Na₂SO₄

These moles of Na₂SO₄ comes from:

2 Na + H₂SO₄ → Na₂SO₄ + H₂

As the reaction is in stoichiometric amounts, moles of Na are twice the moles of Na₂SO₄

0,502 moles of Na₂SO₄ × $\frac{2molesNa}{1moleNa_{2}SO_{4}}$ × 22,99 g/mole = 23,08 g of Na

Molar percent of sodium in original mixture is:

$\frac{23,08g}{26,08g}*100$ = <em>88,50% </em>

I hope it helps

4 0

3 years ago

HELP! ASAP! Iron (Fe) and copper (II) chloride (CuCl2) combine to form iron (III) chloride (FeCl3) and copper (Cu). If you start

Helga [31]

Answer: 30.978

Explanation:

From the equation 2 moles of Fe will result in 3 moles copper

so .325 moles Fe will result in .4875 moles Cu

Cu weights 63.546 gm per mole

.4875 moles * 63.546 gm / mole = 30.978 gm of Cu

7 0

3 years ago

Determine the concentrations of mgcl2, mg2 , and cl– in a solution prepared by dissolving 2.75 × 10–4 g mgcl2 in 1.75 l of water

inysia [295]

Answer:

M of MgCl₂ = 1.65 × 10⁻⁶ M

M of Mg²⁺ = 1.65 × 10⁻⁶ M

M of Cl⁻ = 3.30 × 10⁻⁶ M

Explanation:

1) MgCl₂

Molarity = number of moles of solute / volume of solution in liters, M = n / V

n = mass in grams / molar mass

molar mass of MgCl₂ = 24.305 g/mol + 2(35.543 g/mol) = 95.211 g/mol

n = 2.75 × 10⁻⁴ g / 95.211 g/mol = 2.89×10⁻³ moles

⇒ M = n / V = 2.89×10⁻³ moles / 1.75 l = 1.65 × 10⁻⁶ M

2) Mg²⁺ and Cl⁻

Those are the ions in solution.

You assume 100% dissociation of the ionic compound (strong electrolyte).

Then the equation is: MgCl₂ → Mg²⁺ + 2Cl⁻

That means that 1 mol of MgCl₂ produces 1 mol of Mg²⁺ and 2 moles of Cl⁻.

That yields the same molarity concentration of Mg²⁺ , while the molarity concentration of Cl⁻ is the double.

So, the results are:

M of MgCl₂ = 1.65 × 10⁻⁶ M

M of Mg²⁺ = 1.65 × 10⁻⁶ M

M of Cl⁻ = 3.30 × 10⁻⁶ M

8 0

3 years ago

Read 2 more answers

Which of the following buffers will be most effective at pH 9.25? Group of answer choices a mixture of 1.0 M HC2H3O2 and 1.0 M N

ludmilkaskok [199]

Answer:

The most effective buffer at pH 9.25 will be a mixture of 1.0 M NH3 and 1.0 M NH4Cl

Explanation:

Step 1: Data given

pH of a buffer = pKa + log ([A-]/[Ha])

a mixture of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (Ka for acetic acid = 1.8 x 10-5)

pH = -log( 1.8 * 10^-5) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 4.74

a mixture of 1.0 M NaCN and 1.0 M KCN (Ka for HCN = 4.9 x 10-10)

pH = -log( 4.9 * 10^-10) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 9.30

a mixture of 1.0 M HCl and 1.0 M NaCl

The solution made from NaCl and HCl will NOT act as a buffer.

HCl is a strong acid while NaCl is salt of strong acid and strong base which do not from buffer solutions hence due to HCl PH is less than 7.

a mixture of 1.0 M NH3 and 1.0 M NH4Cl (Kb for ammonia = 1.76 x 10^-5)

Ka * Kb = 1*10^-14

Ka = 10^-14 / 1.76*10^-5

Ka = 5.68*10^-10

pH = -log( 5.68*10^-10) + log (1/1)

pH = -log( 5.68*10^-10)

pH = 9.25

The most effective buffer at pH 9.25 will be a mixture of 1.0 M NH3 and 1.0 M NH4Cl

8 0

3 years ago

The solubility product for barium sulfate is 1.1 × 10–10. calculate the molar solubility of barium sulfate.

Triss [41]

When it comes to equilibrium reactions in chemistry, there are a lot of equilibrium constants that can be used. In the case of solubility, the appropriate one to use is the equilibrium constant of solubility product denotes as Ksp. This is the concentration of products raised to their coefficients. For example,

cC ⇔ aA + bB

Ksp = {[A^a][B^b]}

Now, for the this problem, the reaction is

BaSO₄ ⇔ Ba²⁺ + SO₄²⁻

The reaction is already balanced. Since we don't know the value of Ba²⁺ and SO₄²⁻, let's denote this at x.

1.1 × 10⁻¹⁰ = [x][x] =[x²]
[x] = [Ba²⁺] = [SO₄²⁻] = [BaSO₄] = 1.049 × 10⁻⁵ M

8 0

3 years ago