Convert 380 mmHg to atm.

Mention any two uses of oxygen gas.

Elodia [21]

Answer:

steel, plastics

<h3>Explanation:</h3>

Hope it helps!

3 0

3 years ago

Carbohydrate loading Group of answer choices involves a reduction in the intensity of workouts with a corresponding increase in

SOVA2 [1]

Answer:

the correct option would be:

The group of response options implies a reduction in the intensity of the workouts with a corresponding increase in the percentage of carbohydrate intake for several days before a competition.

Since the carbohydrate load is an increase in glycogen reserves as an energy source accompanied by a decrease in muscle demand. This is often used in high-performance activities, where strict competencies are required.

Although today some professionals do not support that, but rather support a diet with carbohydrates and proteins.

Explanation:

Carbohydrate loading increases glycogen reserves, it is accompanied by a muscle rest plan, without fatigue of muscle fibers.

The purpose of this is to exhaust the muscle fibers in maximum demands such as the competencies, ensuring a necessary energy source that supplies this reaction, for which glycogen reserves are needed.

7 0

3 years ago

How many grams of N2 gas are present in 1.13 L of gas at 2.09 atm and 291 K?

solong [7]

Answer:

Mass= 2.77g

Explanation:

Applying

P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28

PV=nRT

NB

Moles(n) = m/M

PV=m/M×RT

m= PVM/RT

Substitute and Simplify

m= (2.09×1.13×28)/(0.082×291)

m= 2.77g

6 0

3 years ago

First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

Klio2033 [76]

Rate equation for first order reaction is as follows:

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}$

Here, k is rate constant of the reaction, t is time of the reaction, $A_{0}$ is initial concentration and $A_{t}$ is concentration at time t.

The rate constant of the reaction is $0.1 day^{-1}$ .

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days$

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days$

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c) Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days$

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0

2 years ago

(only reply if you know the answer or get reported) Please solve this ty

kolezko [41]

Answer:

7) 50 8)4

Explanation:

I answered the question what do you want to gave me

4 0

2 years ago