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2 years ago

Please help me!!!!!!!!!!!!!

Chemistry

1 answer:

fenix001 [56]2 years ago

4 0

1)The limiting reactant will be aluminum acetate

2) The mass of aluminum hydroxide formed will be 9.75 grams.

<h3>Stoichiometric problem</h3>

The equation of the reaction is as below:

$Al(C_2H_3O_2)_3 + 3NaOH --- > 3C_2H_3NaO_2 + Al(OH)_3$

The mole ratio of the 2 reactants is 1:3.

Mole of 100 mL. 1.25 mol/L $Al(C_2H_3O_2)_3$ = 0.1 x 1.25 = 0.125 mol

Mole of 300 mL, 2.30 mol/L NaOH = 0.3 x 2.3 = 0.69 mol

Thus, aluminum acetate is limited.

Mole ratio of $Al(C_2H_3O_2)_3$ and $Al(OH)_3$ = 1:1

Equivalent mole of $Al(OH)_3$ = 0.125 mol

Mass of 0.125 mole $Al(OH)_3$ = 0.125 x 78 = 9.75 grams.

More on stoichiometric calculations can be found here: brainly.com/question/27287858

#SPJ1

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A mixture of methanol and methyl acetate contains 15.0 wt% methanol.

Usimov [2.4K]

The mixture flow rate in lbm/h = 117.65 lbm/h

<h3>Further explanation</h3>

Given

15.0 wt% methanol

The flow rate of the methyl acetate :100 lbm/h

Required

the mixture flow rate in lbm/h

Solution

mass of methanol(CH₃OH, Mw= 32 kg/kmol) in mixture :

$\tt 15\%\times 200~kg=30~kg\\\\mol=\dfrac{mass}{MW}=\dfrac{30~kg}{32~kg/kmol}=0.9375~kmol$

mass of the methyl acetate(C₃H₆O₂,MW=74 kg/kmol,85% wt) in 200 kg :

$\tt 85\%\times 200=170~kg\\\\mol=\dfrac{170}{74}=2.297~kmol$

Flow rate of the methyl acetate in the mixture is to be 100 lbm/h.

1 kg mixture = 0.85 .methyl acetate

So flow rate for mixture :

$\tt \dfrac{1~kg~mixture}{0.85~methyl~acetat}\times 100~lbm/h=117.65~lbm/h$

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2 years ago

What two things combine with oxygen to form ground-level ozone?

STALIN [3.7K]

Answer:

c) NOx and VOC

oxides of nitrogen (NOx) and volatile organic compounds (VOC)

Explanation:

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3 years ago

- Write down Rutherford's atomic model with diagram .also write down defects of Rutherford's atomic model. - write down neil boh

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3 years ago

What unknown quantity can be calculated after performing a titration? a. volume c. mass b. concentration d. density

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3 years ago

How much water would be needed to completely dissolve 1.52 L of the gas at a pressure of 730 torr and a temperature of 21 ∘C?

Sophie [7]

Answer:

The correct answer is 0.4 L.

Explanation:

The mentioned question can be solved by using the equation,

C = K × Pgas--------(i)

Here K is the Henry law constant whose value is 0.158 mol/L/atm, C is the concentration of the gas in liquid state, and Pgas is the partial pressure of the gas.

Now to find the volume of water, the formula to be used is,

PV = nRT-----------(ii)

Here P is the pressure of the gas, V is the volume, R is the universal gas constant whose value is 0.082 Latm/mol/K and T is the temperature.

PgasV = nRT

Pgas = nRT/Vgas

The value of Pgas is inserted in equation (i) we get,

C = K × nRT/Vgas

It is to be noted that C = n/V, here n is the no. of the moles and V is the volume of liquid.

n/Vliquid = K × nRT/Vgas

1/Vliquid = KRT/Vgas

Vliquid = Vgas/KRT--------------(iii)

Based on the given information, the volume of the gas is 1.52 L, the value of K is 0.158 mol/L/atm, the value of R is 0.082 Latm/mol/K and value of T is 21 degree C or 273 + 21 = 294 K.

Now putting the values in equation (iii) we get,

Vliquid = 1.52 L / 0.158 × 0.082 × 294

Vliquid = 1.52 / 3.809

Vliquid = 0.399 or 0.4 L

Hence, the volume of water required to dissolve 1.52 L of gas is 0.4 L.

8 0

3 years ago