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2 years ago

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The rate law of the reaction NH3 + HOCl → NH2Cl + H2O is rate = k[NH3][HOCl] with k = 5.1 × 106 L/mol·s at 25°C. The reaction is

made pseudo-first-order in NH3 by using a large excess of HOCl. How long will it take for 40% of the NH3 to react if the initial concentration of HOCl is 2 × 10−3 M?

1 answer:

SIZIF [17.4K]2 years ago

6 0

Answer:

40% of the ammonia will take 4.97x10^-5 s to react.

Explanation:

The rate is equal to:

R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]

R = k´ * [NH3]

k´ = 10200 s^-1

Because k´ is the psuedo first-order rate constant, we have the following:

b/(b-x) = 100/(100-40) ; 40% ammonia reacts

b/(b-x) = 1.67

log(b/(b-x)) = log(1.67)

log(b/(b-x)) = 0.22

the time will equal to:

t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s

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In a few sentences, describe the molecular polarity and the intermolecular forces present in ammonium lauryl sulfate.

Sergio [31]

Answer:

A polar molecule is a molecule in which one end of the molecule is slightly positive, while the other end is slightly negative. A diatomic molecule that consists of a polar covalent bond, such as HF, is a polar molecule. The two electrically charged regions on either end of the molecule are called poles, similar to a magnet having a north and a south pole. A molecule with two poles is called a dipole. Hydrogen fluoride is a dipole. A simplified way to depict polar molecules is pictured below When placed between oppositely charged plates, polar molecules orient themselves so that their positive ends are closer to the negative plate and their negative ends are closer to the positive plate

Experimental techniques involving electric fields can be used to determine if a certain substance is composed of polar molecules and to measure the degree of polarity.

For molecules with more than two atoms, the molecular geometry must also be taken into account when determining if the molecule is polar or nonpolar. is a comparison between carbon dioxide and water. Carbon dioxide (CO2) is a linear molecule. The oxygen atoms are more electronegative than the carbon atom, so there are two individual dipoles pointing outward from the C atom to each O atom. However, since the dipoles are of equal strength and are oriented in this way, they cancel each other out, and the overall molecular polarity of CO2 is zero.

Water is a bent molecule because of the two lone pairs on the central oxygen atom. The individual dipoles point from the H atoms toward the O atom. Because of the shape, the dipoles do not cancel each other out, and the water molecule is polar. In the figure, the net dipole is shown in blue and points upward.

Some other molecules are shown below (Figure below). Notice that a tetrahedral molecule such as CH4 is nonpolar. However, if one of the peripheral H atoms is replaced by another atom that has a different electronegativity, the molecule becomes polar. A trigonal planar molecule (BF3) may be nonpolar if all three peripheral atoms are the same, but a trigonal pyramidal molecule (NH3) is polar.

7 0

2 years ago

What pressure (in atm) would be exerted by 76 g of fluorine gas (f2) in a 1.50 liter vessel at -37oc? (a) 26 atm(b) 4.1 atm(c) 1

nirvana33 [79]

<span>Let's assume that the F</span>₂ gas has ideal gas behavior.

<span> Then we can use ideal gas formula,
PV = nRT

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span>⁻¹ K⁻<span>¹) and T is temperature in Kelvin.</span>

Moles = mass / molar mass

Molar mass of F₂ = 38 g/mol

Mass of F₂ = 76 g

Hence, moles of F₂ = 76 g / 38 g/mol = 2 mol

<span>
P = ?
V = 1.5 L = 1.5 x 10</span>⁻³ m³

n = 2 mol

R = 8.314 J mol⁻¹ K⁻<span>¹
T = -37 °C = 236 K

By substitution,
</span>

P x 1.5 x 10⁻³ m³ = 2 mol x 8.314 J mol⁻¹ K⁻¹ x 236 K

p = 2616138.67 Pa

p = 25.8 atm = 26 atm

Hence, the pressure of the gas is 26 atm.

Answer is "a".

<span>

</span>

5 0

2 years ago

The solubility of glucose at 30°C is 125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water a

seraphim [82]

The simplified solubility of glucose at 30°C is 1.25 g/g of water. Considering that the density of water at 30°C is 1 g/mL, the equivalent mass of 400 mL of water is also 400g.

The concentration of the solution in water is,
550 g/400g of water = 1.375 g glucose / g of water

Since the concentration is higher compared to the solubility of glucose at the specified temperature, it can be said that the solution is SATURATED.

4 0

2 years ago

Read 2 more answers

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

8 0

2 years ago

If Mary drove 525miles in 7hours at an average speed of 75 miles per hour how much faster would her average speed need to be to

Makovka662 [10]

<h3>Answer:</h3>

30 miles per hour faster

<h3>Explanation:</h3>

We are given;

Distance that Mary drove as 525 miles
Time she took as 7 hours
Average speed as 75 miles per hour

We are supposed to determine how fast the average speed would be if she made the trip in 5 hours.

We know that speed is given by dividing distance by time taken.
In this case distance remained constant

Therefore, we can determine the speed when she took 5 hours

Speed = Distance ÷ time

= 525 miles ÷ 5 hours

= 105 hours

Thus, the new speed would be 105 miles per hours

But, initial speed was 75 miles per hour

Therefore, she would travel 30 miles per hour faster to cover the journey in 5 hours.

6 0

2 years ago