What is the percent composition of a solution in which 480 grams of sodium chloride, NaCl, is dissolved in 4 liters of solution.

According to the following reaction, how many moles of Fe(OH)2 can form from 175.0 mL of 0.227 M LiOH solution? Assume that ther

IgorC [24]

Answer:

0.020 moles of $Fe(OH)_{2}$ can be formed

Explanation:

1. First determine the number of moles of LiOH.

Molarity is given by the following expression:

$M=\frac{molesofsolute}{Litersofsolution}$

Solving for moles of solute:

moles of solute = M * Liters of solution

Converting 175.0mL to L:

$175.0mL*\frac{1L}{1000mL}=0.175L$

Replacing values:

moles of solute = 0.227M*0.175L

moles of solute = 0.040

Therefore there are 0.040 moles of LiOH

2. Then write the balanced chemical reaction and use the stoichiometry of the reaction to calculate the number of moles of $Fe(OH)_{2}$ produced:

$FeCl_{2}(aq)+2LiOH(aq)=Fe(OH)_{2}(s)+2LiCl(aq)$

As the problem says that there are excess of $FeCl_{2}$ , the limiting reagent is the LiOH.

$0.040molesLiOH*\frac{1molFe(OH)_{2}}{2molesLiOH}=0.020molesFe(OH)_{2}$ can be formed

3 0

2 years ago

(a) At what substrate concentration would an enzyme with a kcat of 30.0 s−1 and a Km of 0.0050 M operate at one-quarter of its m

Dmitrij [34]

The missing graph is in the attachment.

Answer: (a) [S] = 0.0016M

(b) Vmax = 3V; Vmax = $\frac{3V}{2}$ ; Vmax = $\frac{11V}{10}$

(c) Enzyme A: black graph; Enzyme B = red graph

Explanation: Enzyme is a protein-based molecule that speed up the rate of a reaction. Enzyme Kinetics studies the reaction rates of it.

The relationship between substrate and rate of reaction is determined by the Michaelis-Menten Equation:

$V=\frac{V_{max}[S]}{K_{M}+[S]}$

in which:

V is initial velocity of reaction

Vmax is maximum rate of reaction when enzyme's active sites are saturated;

[S] is substrate concentration;

Km is measure of affinity between enzyme and its substrate;

(a) To determine concentration:

$0.25V_{max}=\frac{V_{max}[S]}{0.005+[S]}$

$0.25V_{max}(0.005+[S])=V_{max}[S]$

$0.00125+0.25[S]=[S]$

0.75[S] = 0.00125

[S] = 0.0016M

For a Km of 0.005M, substrate's concentration is 0.0016M.

(b) Still using Michaelis-Menten:

$V=\frac{V_{max}[S]}{K_{M}+[S]}$

Rearraging for Vmax:

$V_{max}=\frac{V(K_{M}+[S])}{[S]}$

(b-I) for [S] = 1/2Km

$V_{max}=\frac{V(K_{M}+0.5K_{M})}{0.5K_{M}}$

$V_{max}=\frac{V(1.5K_{M})}{0.5K_{M}}$

$V_{max}=$ 3V

(b-II) for [S] = 2Km

$V_{max}=\frac{V(K_{M}+2K_{M})}{2K_{M}}$

$V_{max}=\frac{V(3K_M)}{2K_M}$

$V_{max}=\frac{3V}{2}$

(b-III) for [S] = 10Km

$V_{max}=\frac{V(K_{M}+10K_M)}{10K_M}$

$V_{max}=\frac{V(11K_{M})}{10K_{M}}$

$V_{max}=\frac{11V}{10}$

(c) Being the affinity between enzyme and substrate, the lower Km is the less substrate is needed to reach half of maximum velocity.

Km of enzyme A is 2μM and of enzyme B is 0.5μM.

Enzyme B has lower Km than enzyme A, which means the first will need a lower concnetration of substrate to reach half of Vmax.

Analyzing each plot, notice that the red-coloured graph reaches half at a lower concentration, therefore, red-coloured plot is for enzyme B, while black-coloured plot is for enzyme A

3 0

1 year ago

What mass of oxygen reacts when 84.9 g of iron is consumed in the following reaction: Fe+O2= Fe2O3

konstantin123 [22]

First we will calculate the number of moles of Iron:
$n = \frac{m}{M}
$ , where n is the number of moles, m is the mass of iron in the reaction and M is the Atomic weight.
$n= \frac{84.9}{55.845} = 1,52$ moles of Iron.
The same number of moles of Oxygen will take part in the reaction.
So $1,52= \frac{mOxygen}{32}$ where 32 is the Atomical Weight of Oxygen (16 x 2).
=> $mOxygen=32*1,52=48,64$ g

6 0

1 year ago

Read 2 more answers

How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mm hg and a temperature of 21.0°c, can be made

natta225 [31]

Answer:
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.
1.566 g Zn x (1 mole Zn / 65.38 g Zn) = 0.02395 moles Zn
0.02395 moles Zn x (1 mole H2 / 1 mole Zn) = 0.02395 moles H2 produced
Now use the ideal gas law to find the volume V.
P = 733 mmHg x (1 atm / 760 atm) = 0.964 atm
T = 21 C + 273 = 294 K
PV = nRT
V = nRT/ P = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm) = 0.600 L

7 0

2 years ago

The density of an unknown crystal is 2.65gmL. If a 4.46 g sample of the crystal is added to a graduated cylinder containing 25.0

jolli1 [7]

Answer:

Total volume after adding crystal = 26.7 mL

Explanation:

Given data:

Density of crystal = 2.65 g/mL

Mass of sample = 4.46 g

Volume of water = 25.0 mL

Volume after adding crystal = ?

Solution:

First of all we will calculate the volume of crystal.

d = m/v

2.65 g/mL = 4.46 g/ v

v = 4.46 g/2.65 g/mL

v = 1.7 mL

Total volume after adding crystal = Volume of water + Volume of metal

Total volume after adding crystal = 25.0 mL + 1.7 mL

Total volume after adding crystal = 26.7 mL

8 0

1 year ago