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Ivenika [448]
3 years ago
14

Which statement best describes how a catalyst increases the rate of a reaction?

Chemistry
2 answers:
Reika [66]3 years ago
6 0

Answer:

<h2>b</h2>

Explanation:

Ainat [17]3 years ago
3 0
A catalyst "provides an alternate reaction pathway with a lower activation energy". (2) is the correct answer.

Activation energy is the energy needed to conduct a chemical reaction. When a catalyst is put into the system, it makes the system require less activation energy, making the reaction go faster.

Hope this is helpful~ 
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Looking at page 14 of the earth science tables, how does the temperature change as a balloon rises through the Troposphere Strat
givi [52]

Answer:

Decreases

Explanation:

The temperature decreases when latitude increases.

5 0
3 years ago
Which of the following molecules has correctly labeled bond dipoles?
Pepsi [2]
The bond dipole moment<span> uses the idea of </span>electric dipole moment<span> to measure the </span>polarity<span> of a chemical bond within a </span>molecule<span>. It occurs whenever there is a separation of positive and negative charges. In the diagram above, option B exhibited a bond dipole moment. I hope this helps.</span>
8 0
3 years ago
How many grams are in 1 mole of Ar?
Ilia_Sergeevich [38]

Answer: 39.948 grams

Explanation:

The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Ar, or 39.948 grams

3 0
3 years ago
NEED HELP ASAP
katrin [286]
Removing seed casings from grains is SEPARATING. a soda bubble bubbling when it is opened is MIXING. a bright copper statue turning green is MIXING. remove salt from seawater is SEPARATING. water decomposing is SEPARATING.
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
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