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3 years ago

SEP Develop Models Each helium atom has two protons. Sketch models

Chemistry

1 answer:

RSB [31]3 years ago

6 0

Protons and neutrons are located in the nucleus of the atom while the electrons move in the trajectory of the shell

<h3>Further explanation </h3>

Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

Isotopes of Helium : helium-3 and helium-4

₂³He :

protons = 2

electrons=protons=2

neutron=mass number-atomic number=3-2=1

₂³He :

protons = 2

electrons=protons=2

neutron=mass number-atomic number=3-2=1

₂⁴He

protons = 2

electrons=protons=2

neutron=mass number-atomic number=4-2=2

Protons and neutron in the nucleus, electrons in the shell

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1. Which phase change is accompanied by the release of heat? A) H20(s)--> H2O(g) B) H20(l) ->H2O(s) C) H20(l)→ H2O(g) D) H

docker41 [41]

Answer:

B, liquid to solid.

Explanation: Since heat is being released, the particles for H2O would clump up. Heat is basically being taken out.

5 0

3 years ago

A serving of Cheez-Its releases 1.30 x 10^4 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy we

kondor19780726 [428]

Answer:

The final temperature is:- 7428571463.57 °C

Explanation:

The expression for the calculation of heat is shown below as:-

$Q=m\times C\times \Delta T$

Where,

$Q$ is the heat absorbed/released

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of water = 1.75 mg = 0.00175 g ( 1 g = 0.001 mg)

Specific heat of water = 4.18 J/g°C

Initial temperature = 35 °C

Final temperature = x °C

$\Delta T=(x-35)\ ^0C/tex]
Q = [tex]1.3\times 10^4$ kcal

Also, 1 kcal = 4.18 kJ = $4.18\times 10^3$ J

So, Q = $1.3\times 10^4\times 4.18\times 10^3$ J = 54340000 J

So,

$54340000=0.00175\times 4.18\times (x-35)$

$0.00175\times \:4.18\left(x-35\right)=54340000$

$x-35=\frac{54340000}{0.007315}$

$x=7428571463.57$

Thus, the final temperature is:- 7428571463.57 °C

3 0

3 years ago

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.

timofeeve [1]

Complete question:

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

Answer:

The magnitude of q for the process 568 J.

Explanation:

Given;

change in internal energy of the gas, ΔU = 475 J

work done by the gas, w = 93 J

heat added to the system, = q

During gas expansion process, heat is added to the gas.

Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

ΔU = q - w

q = ΔU + w

q = 475 J + 93 J

q = 568 J

Therefore, the magnitude of q for the process 568 J.

6 0

3 years ago

Balanced chemical equations show:

lara31 [8.8K]

the products formed from the reaction

all of the above

4 0

3 years ago

ANSWER FAST PLZZZ!!!!!!!!!!!!!

goldfiish [28.3K]

Answer:

1 and 2

Explanation:

1 is negative

2 is negative

3 is positive

5 0

3 years ago