Ask question

Ask question

All categories

3 years ago

12

Please help me i need yall will mark brainliest and 30 points!

1 answer:

Amiraneli [1.4K]3 years ago

8 0

Reaction

2Fe(OH)₃ + 3H₂SO₄⇒Fe₂(SO₄)₃ + 6H₂O

because the limiting reactant is Fe(OH)₃, then the moles of all compounds will be based on moles of Fe(OH)₃

mole H₂SO₄ = 3/2 x mole Fe(OH)₃ = 3/2 x 3 mole = 4.5 mole

remaining mole of H₂SO₄ (as an excess reactant) = 6.4 - 4.5 = 1.9 mole

You might be interested in

Penicillinase, also known as β‑lactamase, is a bacterial enzyme that hydrolyzes and inactivates the antibiotic penicillin. Penic

sergeinik [125]

Explanation:

The given data is as follows.

$V_{max} = 6.8 \times 10^{-10} \mu mol/min$

$K_{m} = 5.2 \times 10^{-6} M$

Now, according to Michaelis-Menten kinetics,

$V_{o} = V_{max} \times [\frac{S}{(S + Km)}]$

where, S = substrate concentration = $10.4 \times 10^{-6}$ M

Now, putting the given values into the above formula as follows.

$V_{o} = V_{max} \times [\frac{S}{(S + Km)}]$

$V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]$

$V_{o} = 6.8 \times 10^{-10} \mu mol/min \times 0.667$

= $4.5 \times 10^{-10} \mu mol/min$

This means that $V_{o}$ would approache $V_{max}$ .

5 0

4 years ago

Which feature is used to classify galaxies? age color shape size

kondor19780726 [428]

Answer:

C. Shape

Explanation:

Did the quiz! & galaxies are classified by shapes called "elliptical, spirial, & irregular"

Good luck :)

8 0

3 years ago

The Earth's period of revolution is 365.25 days. Convert this period of revolution into hours

Helga [31]

8,766 hours is your answer :)

4 0

3 years ago

Breve resumen de que es volumen

andrey2020 [161]

What??? what language is this?

4 0

3 years ago

Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

Mass of water = m= 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute(NaCl)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

Mass of water = m = 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p'$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute ( glucose)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

3 0

3 years ago