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TiliK225 [7]
3 years ago
5

Where did coal originate ?​

Chemistry
2 answers:
liubo4ka [24]3 years ago
8 0

Answer:

Coal is a fossil fuel and is the altered remains of prehistoric vegetation that originally accumulated in swamps and peat bogs

g100num [7]3 years ago
4 0

Coal is a fossil fuel and is the altered remains of prehistoric vegetation that originally accumulated in swamps and peat bogs.
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There are two naturally occurring isotopes of boron. 10 B has a mass of 10.0129 u. 11 B has a mass of 11.0093 u. Determine the a
Vanyuwa [196]
<h2>Natural Abundance for 10B is 19.60%</h2>

Explanation:

  • The natural isotopic abundance of 10B is 19.60%.
  • The natural isotopic abundance of 11B is 80.40%.
  • The isotopic masses of boron are 10.0129 u and 11.009 u respectively.

For calculation of abundance of both the isotopes -

Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.

Determining it as an equation -

10x + 11y= 10.8

x+y=1 (ratio)

10x + 10y = 10

By taking the denominator away from the numerator

we get;

y = 0.8

x + y = 1

∴ x = 0.2

To get percentages  we need to multiply it by 100

So, the calculated abundance is 80% for 11 B and 20% 10  B.

5 0
3 years ago
Sometimes voltage is also called A. electrical force. B. electrical potential. C. electrical resistance. D. electrical field ene
Oliga [24]
Voltage<span>, </span>also called<span> electromotive force, is a quantitative expression of the potential difference in charge between two points in an electrical field.

So ACTUALLY an "electromotive force", but of your answer choices.

D. Electrical Field Energy
</span>
7 0
3 years ago
Read 2 more answers
¿sabes cual es la concentración de azúcar en un refresco de un litro?​
Grace [21]
“Esto quiere decir que en una lata de 330 ml tiene alrededor de 36 gramos de azúcar, mientras que una botella de 1 l tendrá 110 gramos”

110 gramos
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
3 years ago
Light rays enter a translucent material what happening to the light rays
artcher [175]

When this happens, light changes speed and the light ray bends, either toward or away from what we call the normal line, an imaginary straight line that runs perpendicular to the surface of the object.

6 0
3 years ago
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