There would be an equal amounts of ELECTRONS.

In order for the atom to be neutral it would imply that the number of protons (positive particles) and the number of electrons (negative particles) are equal since the neutrons are without charge.

3 0

2 years ago

Exactly 10.0 L of air -25°C is heated to 100.0°C. What is the new volume if the pressure is kept constant?

zepelin [54]

Answer: V2= 15.0403226 Liters

Explanation:

Use V1/T1=V2/T2

Make sure you change the degrees Celsius to Kelvin. (Kelvin = degrees Celsius +273)

10.0L / 248 K = V2/ 373 K

Cross multiply V1 and T2 and divide by T1

(10.0 L)( 373K)/ 248 K = V2

V2= 15.0403226 Liters (Kelvin cancels out)

4 0

2 years ago

) Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2is 7

Travka [436]

Answer:

Rate of the reaction is 0.2593 M/s

-0.5186 M/s is the rate of the loss of ozone.

Explanation:

The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.

$2O_3\rightleftharpoons 3O_2$

Rate of formation of oxygen : $7.78\times 10^{-1} M/s$

Rate of the reaction(R) = $\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}$

$R=\frac{1}{3}\frac{d[O_2]}{dt}$

Rate of formation of oxygen=3 × (R)

$7.78\times 10^{-1} M/s=3\times R$

Rate of the reaction(R): $0.2593 M/s$

Rate of the reaction is 0.2593 M/s

Rate of disappearance of the ozone:

$R=-\frac{1}{2}\frac{d[O_3]}{dt}$

$\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s$

-0.5186 M/s is the rate of the loss of ozone.

6 0

2 years ago

Which organelles surround the cell?

GalinKa [24]

Answer:

Cell Wall, Cell Membrane, Nucleus, Endoplasmic Reticulum (ER), and Ribosome.

Explanation:

Hopefully this helps :).

5 0

2 years ago

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