All categories

3 years ago

If the protons are 6 the neutrons are 6 and the electrons are 6 what is the total charge

Chemistry

1 answer:

horrorfan [7]3 years ago

6 0

Answer:

General Formulas and Concepts:

Chemistry - Atomic Structure

Protons have a positive (+) charge.
Neutrons have a neutral (0) charge.
Electrons have a negative (-) charge.

Explanation:

Step 1: Define

6 Protons

6 Neutrons

6 Electrons

Step 2: Determine charges

6 Protons = + 6 charge

6 Neutrons = 0 charge

6 Electrons = - 6 charge

Step 3: Determine overall charge

+6 + 0 + -6 = 0

You might be interested in

How is reusable water bottles are more environmentally friendly than one use water bottle?

dexar [7]

Answer:

They are better because they are more usable than Plastic and reusable water bottles are better for the planet!

Explanation:

6 0

3 years ago

How many liters of water vapor are in 36.21 g?

Elden [556K]

Answer:

45.02 L.

Explanation:

Firstly, we need to calculate the no. of moles of water vapor.

n = mass / molar mass = (36.21 g) / (18.0 g/mol) = 2.01 mol.

We can calculate the volume of knowing that 1.0 mole of a gas at STP occupies 22.4 L.

Using cross multiplication:

1.0 mole of CO occupies → 22.4 L.

2.01 mole of CO occupies → ??? L.

∴ The volume of water vapor in 36.21 g = (22.4 L)(2.01 mole) / (1.0 mole) = 45.02 L.

4 0

3 years ago

The temperature of a sample of water increases from 20c to 46.6c as it absorbs 5650 Joules of heat. What is the mass of the samp

WINSTONCH [101]

Answer: 51 g

Explanation:

5 0

2 years ago

A 5 grams sample of liquid water evaporates into gaseous water in a closed container, expanding in the process. What is the mass

padilas [110]

We have get the mass of gaseous water after evaporation in a closed container.

The mass of water vapor after evaporation is 5 grams.

In closed container, there is no exchange in mass from system to surrounding, only heat may exchange. The number of moles of water vapour remains unchanged as 5 gram water is heated in closed container.

Due to heating, liquid water gets evaporated and intermolecular distance between water molecules increases in gaseous state than liquid state and intermolecuar force of attraction decreases.

Randomness of molecules increases in gaseous state than liquid state.

5 0

3 years ago

A 3.00-kg block of copper at 23.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen

hammer [34]

Answer:

1.2584kg of nitrogen boils.

Explanation:

Consider the energy balance for the overall process. There are not heat or work fluxes to the system, so the total energy keeps the same.

For the explanation, the 1 and 2 subscripts will mean initial and final state, and C and N2 superscripts will mean copper and nitrogen respectively; also, liq and vap will mean liquid and vapor phase respectively.

The overall energy balance for the whole system is:

$U_1=U_2$

The state 1 is just composed by two phases, the solid copper and the liquid nitrogen, so: $U_1=U_1^C+U_1^{N_2}$

The state 2 is, by the other hand, composed by three phases, solid copper, liquid nitrogen and vapor nitrogen, so:

$U_2=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

So, the overall energy balance is:

$U_1^C+U_1^{N_2}=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

Reorganizing,

$U_1^C-U_2^C=U_{2,liq}^{N_2}+U_{2,vap}^{N_2}-U_1^{N_2}$

The left part of the equation can be written in terms of the copper Cp because for solids and liquids Cp≅Cv. The right part of the equation is written in terms of masses and specific internal energy:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_1^{N_2}u_1^{N_2}$

Take in mind that, for the mass balance for nitrogen, $m_1^{N_2}=m_{2,liq}^{N_2}+m_{2,vap}^{N_2}$ ,

So, let's replace $m_1^{N_2}$ in the energy balance:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,liq}^{N_2}u_1^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

So, as you can see, the term $m_{2,liq}^{N_2}u_{2,liq}^{N_2}$ disappear because $u_{2,liq}^{N_2}=u_{1,liq}^{N_2}$ (The specific energy in the liquid is the same because the temperature does not change).

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}(u_{2,vap}^{N_2}-u_1^{N_2})$

The difference $(u_{2,vap}^{N_2}-u_1^{N_2})$ is the latent heat of vaporization because is the specific energy difference between the vapor and the liquid phases, so:

$m_{2,vap}^{N_2}=\frac{m_C*Cp*(T_1^C-T_2^C)}{(u_{2,vap}^{N_2}-u_1^{N_2})}$

$m_{2,vap}^{N_2}=\frac{3kg*0.092\frac{cal}{gC} *(296.15K-77.3K)}{48.0\frac{cal}{g}}\\m_{2,vap}^{N_2}=1.2584kg$

3 0

3 years ago