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Degger [83]
3 years ago
10

Which of the following processes is responsible for producing large amounts of snow in Buffalo, New York? A.) mountain lift effe

ct B.) lake effect C.) urban canyon effect D.) urban heat island
Chemistry
2 answers:
adell [148]3 years ago
8 0
The appropriate answer is b. the lake effect. Buffalo is located on the north eastern shore of the Lake Erie. The high snowfall in this area is due to the fact that cold air masses will move across the expanse of the warmer lake water during the winter months. This warms the layer closest to the water causing it to pick up moisture from the lake. Once the moisture rises through the air mass it will freeze and form snow.
Naddik [55]3 years ago
7 0
B is the correct answer 
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1,4-Pentadiene has a AHhydro = -254 kJ/mol while trans-1,3-pentadiene has a AHhydra = -226 kJ/mol. Explain this difference in he
julsineya [31]

Answer:

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Explanation:

Here, \Delta H_{hydro}=H(hydrogenated pdt.)-H(diene)

H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation

H(diene) depends on stability of diene.

More stable a diene, lesser will be it's H(diene) value (more neagtive).

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Hence, \Delta H_{hydro} is higher (less negative) for trans-1,3-pentadiene

5 0
3 years ago
Xbox or ps4? haha have a. Good ond
bekas [8.4K]

Answer:

I play none but If I did I would choose Xbox

btw, thank you

7 0
3 years ago
5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0
3 years ago
The molecules of a substance must have what characteristics in order to be flexible
Goryan [66]
Answer:
            <span>The molecules of a substance which must have the <u>a</u></span><u>bility to move past one another</u> are said to be flexible.

Explanation:
                   Those substances are said to be flexible which can be bent without breaking. There are many substances which are hard in nature but still can be bent. The hardness of such materials is due to strong interactions between the molecules and the flexibility comes due to their amorphous backbone. Therefore, greater the crystalline level of macromolecules lesser is the flexibility and greater the amorphous character greater is the flexibility and vice versa. Also, the flexibility of polymers is increased by adding plastisizers in it. Plastisizers make the hard polymers flexible by breaking the crosslinkers and enabling the macromolecules to move past one another.

6 0
2 years ago
If the volume of a gas container at 32 degrees Celsius changes from 1.55 L to 755 mL, what will the final temperature be?
QveST [7]
So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
4 0
2 years ago
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