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3 years ago

What is a warm front

Chemistry

2 answers:

Olegator [25]3 years ago

5 0

The changeover region where a warm air mass is replacing a cold air mass

Galina-37 [17]3 years ago

3 0

Answer:the Boundary of an advancing mass of warm air, in particular the leading edge of the warm sector of a low-pressure system.

Explanation:

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The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 625°C th

Alona [7]

Answer:

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_2$ = rate of reaction at $T_2$

$K_1$ = rate of reaction at $T_1$

$Ea$ = activation energy of the reaction

R = gas constant = 8.314 J/K mol

$E_a=300 kJ/mol=300,000 J/mol$

$T_2=625^oC=898.15 K,T_1=525^oC=798.15 K$

$\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]$

$\log (\frac{K_2}{K_1})=2.185666$

$K_2=153.344\times K_1$

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

4 0

3 years ago

Metric prefixes 9th grade level

zhannawk [14.2K]

Answer:

<h3>A metric prefix is a unit prefix that precedes a basic unit of measure to indicate a multiple or submultiple of the unit. All metric prefixes used today are decadic. Each prefix has a unique symbol that is prepended to any unit symbol.</h3>

6 0

3 years ago

Caleb has a 860-kilogram car sitting in his back yard, gathering rust.

emmasim [6.3K]

<h3>Answer:</h3>

A. 860 kg

<h3>Explanation:</h3>

To answer the question we need to understand that;

Mass refers to the amount of matter in an object.
Weight, on the other hand, refers to the gravitational pull of an object to a given surface.
Mass is measured using a spring balance.

We also need to know that;

The mass of an object remains constant every where irrespective of the gravitational acceleration.

Therefore, an object on the surface of the earth would have the same mass as on the surface of the moon.
In this case; the mass of the car remains the same on the outer space as on the back yard.

3 0

3 years ago

Read 2 more answers

1 A 3.80 g sample of bronze was dissolved in sulphuric acid. The copper in alloy reacted with

s344n2d4d5 [400]

Answer:

yeyeye

Explanation:

hi and bye no problem you will come and die

3 0

3 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

3 years ago