A scientist dissolves 10 g of NaOH into 0.275 L of water. What is the

if 334.6 g of phosphoric acid is reacted with excess potassium hydroxide. the final mass K3PO4 produced is found to be 248g. wha

Orlov [11]

Answer:

%age Yield = 34.21 %

Explanation:

The balance chemical equation for the decomposition of KClO₃ is as follow;

3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O

Step 1: Calculate moles of H₃PO₄ as;

Moles = Mass / M/Mass

Moles = 334.6 g / 97.99 g/mol

Moles = 3.414 moles

Step 2: Find moles of K₃PO₄ as;

According to equation,

1 moles of H₃PO₄ produces = 1 moles of K₃PO₄

So,

3.414 moles of H₃PO₄ will produce = X moles of K₃PO₄

Solving for X,

X = 1 mol × 3.414 mol / 1 mol

X = 3.414 mol of K₃PO₄

Step 3: Calculate Theoretical yield of K₃PO₄ as,

Mass = Moles × M.Mass

Mass = 3.414 mol × 212.26 g/mol

Mass = 724.79 g of K₃PO₄

Also,

%age Yield = Actual Yield / Theoretical Yield × 100

%age Yield = 248 g / 724.79 × 100

%age Yield = 34.21 %

3 0

3 years ago

Balance this chemical equation. A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coeffici

larisa86 [58]

The first one have a nice day and good luck

8 0

3 years ago

1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?

RSB [31]

Answer : The number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

$PV=nRT$

where,

P = Pressure of $N_2$ gas = $1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm$ (1 atm = 760 mmHg)

V = Volume of $N_2$ gas = 985 mL = 0.982 L (1 L = 1000 mL)

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $0.0^oC=273+0.0=273K$

Now put all the given values in above equation, we get:

$(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K$

$n=5.78\times 10^{-11}mol$

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains $6.022\times 10^{23}$ number of molecules.

As, 1 mole of $N_2$ gas contains $6.022\times 10^{23}$ number of molecules

So, $5.78\times 10^{-11}$ mole of $N_2$ gas contains $(5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13}$ number of molecules

Therefore, the number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

8 0

3 years ago

Please answer truthfully:))

makvit [3.9K]

Answer:

Fast, Direction, Time

Explanation:

4 0

3 years ago

When a 17.7 mL sample of a 0.368 M aqueous hypochlorous acid solution is titrated with a 0.301 M aqueous barium hydroxide soluti

nordsb [41]

Answer:

pH = 12.98

Explanation:

Step 1: Data given

Volume of aqueous hypochlorous acid solution = 17.7 mL = 0.0177 L

Molarity of aqueous hypochlorous acid solution = 0.368 M

Molarity of aqueous barium hydroxide solution = 0.301 M

Volume of aqueous barium hydroxide solution = 16.2 mL = 0.0162 L

Step 2: The balanced equation

2HCl + Ba(OH)2 → BaCl2 + 2H2O

Step 3: Calculate moles

Moles = molarity * volume

Moles HCl = 0368 M * 0.0177 L

Moles HCl = 0.0065136 moles

Moles Ba(OH)2 = 0.301 M * 0.0162 L

Moles Ba(OH)2 = 0.0048762 moles

Step 4: Calculate the limiting reactant

For 2 moles HCl we need 1 mol Ba(OH)2 to produce 1 mol BaCl2 and 2 moles H2O

HCl is the limiting reactant. It will completely be consumed 0.0065136 moles. Ba(OH)2 is in excess. There will react 0.0065136/2 = 0.0032568‬ moles. There will remain 0.0048762 moles - 0.0032568‬ = 0.0016194 moles

Step 5: Calculate molarity Ba(OH)2

Molarity Ba(OH)2 = moles / volume

Molarity Ba(OH)2 = 0.0016194 moles / 0.0339 L

Molarity Ba(OH)2 = 0.04777 M

Step 6: Calculate [OH-]

Ba(OH)2 → Ba^2+ + 2OH-

For Ba(OH)2 we have 2* [OH-]

[OH-] = 2*0.04777 = 0.09554 M

Step 7: Calculate pOH

pOH = -log[OH-]

pOH = -log(0.09554)

pOH = 1.02

Step 8: Calculate pH

pH = 14 - 1.02

pH = 12.98

8 0

3 years ago