How many molecules are there in 79g of Fe2O3? how many atoms is this?

If 0.640 g of beautiful blue crystals of azulene is dissolve in 99 g of benzen, the resulting solutions boils at 80.23 degrees c

devlian [24]

This problem handles boiling-point elevation, which means we will use the formula:

ΔT = Kb * m

Where ΔT is the difference of Temperature between boiling points of the solution and the pure solvent (Tsolution - Tsolvent). Kb is the ebullioscopic constant of the solvent (2.64 for benzene), and m is the molality of the solution.

Knowing that benzene's boiling point is 80.1°C, we solve for m:

Tsolution - Tsolvent = Kb * m

80.23 - 80.1 = 2.64 * m

m = 0.049 m

We use the definition of molality to calculate the moles of azulene:

0.049 m = Xmoles azulene / 0.099 kgBenzene

Xmoles azulene = 4.87 x10⁻³ moles azulene

We use the mass and the moles of azulene to calculate its molecular weight:

0.640 g / 4.875 x10⁻³ mol = 130.28 g/mol

A molecular formula that would fulfill that molecular weight is C₁₀H₁₀. So that's the result of solving this problem.

The actual molecular formula of azulene is C₁₀H₈.

6 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

2 years ago

36g of KOH dissolved in 800mL of water. What is the molality of the solution?

Bad White [126]

Answer:

0.80m of KOH

Explanation:

Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.

In the problem, the solute is KOH and solvent is water.

Moles of 36g KOH -Molar mass: 56.1g/mol- are:

36g KOH × (1mol / 56.1g) = 0.642 moles of KOH

Now, as density of water is 1g/mL, mass of 800mL of water is:

800mL × (1g / mL) × (1kg / 1000g) = 0.800kg of water

Thus, molality is:

0.642moles of KOH / 0.800kg = 0.80m of KOH

5 0

2 years ago

Item 5 A solution of methanol, CH3OH, in water is prepared by mixing together 128 g of methanol and 108 g of water. The mole fra

Basile [38]

Answer:

Mole fraction of methanol will be closest to 4.

Explanation:

Given, Mass of methanol = 128 g

Molar mass of methanol = 32.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{128\ g}{32.04\ g/mol}$

$Moles\ of\ methanol = 3.995\ mol$

Given, Mass of water = 108 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{108\ g}{18.0153\ g/mol}$

$Moles\ of\ water= 5.995\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ methanol=\frac {n_{methanol}}{n_{methanol}+n_{water}}$

$Mole\ fraction\ of\ methanol=\frac{3.995}{3.995+5.995}=0.39989$

Mole fraction of methanol will be closest to 4.

5 0

2 years ago

Which of these statements is not necessarily true for two objects in thermal equilibrium?

Dmitrij [34]

1. Answer: C. The objects' temperatures have both changed by the same amount.

Explanation:

An object is said to be in thermal equilibrium when the objects have attained same temperature. Heat transfer from hotter object to colder one in contact takes place until the temperature of the two are equal. It is not necessary that the temperature of both the objects changes by same amount. After attainment of thermal equilibrium, the temperature of the objects stop changing and the tiny particles of the object move at the same rate.

Hence, the objects' temperatures have both changed by the same amount. is not necessarily true for two objects in thermal equilibrium.

2. Answer: C. Objects are made of tiny particles, and their motion depends on the temperature.

Explanation:

Kinetic theory of heat states that the kinetic energy of constituent particles determine the temperature of the object. The statement that best explains this is Objects are made of tiny particles, and their motion depends on the temperature.

5 0

2 years ago