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Which of the following will affect the rate of a chemical reaction?

dimulka [17.4K]

Answer:

Solute concentration will afect the rate of a chemical reaction, because you must work with molarity

Explanation:

I think that solute mass may be it can affect the rate of reaction, if you have more mass in a solute, you will also have more moles.

If you want to know more, you have to consider temperature in the reaction and the presence of catalysts. They all, affect reactions.

7 0

2 years ago

Read 2 more answers

A chemist conducted a study on ammonia NH3(g) into hydrogen gas and nitrogen gas. This reaction is represented by the following

Lelu [443]

Answer:

.

Explanation:

7 0

2 years ago

What role does the activation energy play in chemical reactions?

Tatiana [17]

The activation energy is the minimum amount of energy that particles must have in order for them to participate in a chemical reaction. During chemical reactions bonds are broken and formed. Particles must collide with sufficient energy in order for the initial bonds to be broken. The activation energy is that that initial minimum energy that the particles can have in order for the bonds to be broken. Stronger bonds would require more energy to be broken and therefore the activation energy for such would be higher.

7 0

3 years ago

What type of bacteria convert the

Ksju [112]

Answer: B

Nitrifying bacterium, plural Nitrifying Bacteria, any of a small group of aerobic bacteria (family Nitrobacteraceae) that use inorganic chemicals as an energy source.

3 0

2 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago