What do you call a substance that influences the reaction but does not participate in the reaction?

1 answer:

8 0

A substance that influences the reaction but does not participate in the reaction is called a catalyst. In other words, it increases the rate of a chemical reaction without itself undergoing any permanent chemical change.

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Name three complex molecules synthesized by plants

DiKsa [7]

Oxygen, glucose and starch
Oxygen and glucose are produced during photosynthesis. Glucose is then stored as starch.

Hope it helped!

3 0

3 years ago

Three positive charges lie on the x axis: q1 = 1 × 10−8 C at x1 = 1 cm, q2 = 2 × 10−8 C at x2 = 2 cm, and q3 = 3 × 10−8 C at x3

NARA [144]

Answer: Option (4) is the correct answer.

Explanation:

Relation between potential energy and charge is as follows.

U = $\frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]$

As it is given that $q_{1} = 1 \times 10^{-8} C$ , $q_{2} = 2 \times 10^{-8} C$ , and $q_{3} = 3 \times 10^{-8} C$ .

Distance between the charges = 1 cm = $1 \times 10^{-2} m$ (as 1 cm = 0.01 m)

Hence, putting these given values into the above formula as follows.

U = $\frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]$

= $9 \times 10^{9} [\frac{1 \times 10^{-8} \times 2 \times 10^{-8}}{10^{-2}} + \frac{2 \times 10^{-8} \times 3 \times 10^{-8}}{10^{-2}} + \frac{3 \times 10^{-8} \times 1 \times 10^{-8}}{10^{-2}}]$

= $9 \times 10^{9} [2 + 6 + 1.5]$

= $85.5 \times 10^{-5}$ J

= 0.00085 J

Thus, we can conclude that the potential energy of this arrangement, relative to the potential energy for infinite separation, is about 0.00085 J.

8 0

3 years ago

Daniel claims that the product of two square roots of integers is always rational. His reasoning is given in the example below.

Katen [24]

Incorrect....................

6 0

4 years ago

The volume of a gas is 650. mL at 1.14°C. What will the volume be a 75.0°C?

tia_tia [17]

Answer: The final volume of the gas is 825.12 mL

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$