What number of moles of O2 will be produced by the decomposition of 4.4 moles of water?

A 100-w light bulb is about 16% efficient. how many joules of heat energy are siddipated by the bulb each second

pickupchik [31]

Answer:

84 Joules are wasted per second.

Explanation:

An efficiency of 16% means 84% is wasted as heat energy.

Therefore;

For a 100W bulb;

Heat wasted = 84/100 × 100 W

= 84 W

Therefore;

84 W or 84 joules are wasted as heat per second

3 0

4 years ago

Need help please help me

Mademuasel [1]

Answer:

The answer is most likely compound a

Explanation:

Covalent compounds have low melting and boiling points and conduct electricity poorly

8 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

4 years ago

Please help

Pachacha [2.7K]

1. C : Mg(CN)2
2. B : N2O5
3. D : Ti(ClO4)3
4. A : Ni(NO3)3
5. D : Cobalt (III) Acetate
6. B : Nickel(II) sulfate
7. C : Dinitrogen Tetrafluoride
8. A : Phosphorus pentachloride<em />
9. C : Metallic <em>(<!> This is the only one I'm not 100% sure of)</em>
10. A : Ionic
11. C : Metallic
12. B : Covalent

5 0

3 years ago

Read 2 more answers

Explain why a coin is able to float on top of water in a glass.

AlexFokin [52]

It is the concept of surface tension that helps the coin to float on the top of the water in the glass. The surface tension acts just opposite to the gravitational force and that is why when any metal placed in the water it will float and come on the surface. Hope this helps!

8 0

4 years ago

Read 2 more answers