Determine which reaction has the highest activation energy.

Which one is NOT part of the cell theory?

worty [1.4K]

<u>Answer:</u>

All living things are not made of cells.

5 0

3 years ago

An element has an atomic number of 30. How many protons and electrons are in a neutral atom of the element? A. 30 protons, 30 el

Minchanka [31]

Atomic number shows the number of protons in an atom's nucleus. If the atomic number is 30, then it has 30 protons. The question says it is neutral, so it must have the same amount of protons and electrons.

A is the correct answer based on this.

7 0

3 years ago

William measures a test tube and finds that the mass of the test tube is 5g. He places the lone reactant in the test tube and fi

iogann1982 [59]

86 percent is the percent yield for this experiment if he expected to produce 5g of product.

Explanation:

Given that:

mass of test tube = 5 grams

mass of test tube + reactant is 12.5 grams

mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)

mass of reactant = 12.5 -5

= 7.5 grams

when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.

so mass of product formed = 9.3 - 5

= 4. 3 grams of product is formed (actual yield)

However, he expected the product to be 5 grams (theoretical yield)

Percent yield = $\frac{actual yield}{theoretical yield}$ x 100

putting the values in the formula:

percent yield = $\frac{4.3}{5}$ x 100

= 86 %

86 percent is the percent yield.

5 0

3 years ago

Please help me ASAP the question is done below

mixas84 [53]

Answer:

Explanation:

First Question. Answer is: B. They are different.

For examle, balanced chemical reaction of forming water from hydrogen and oxygen:

2H₂ + O₂ → 2H₂O.

During chemical reaction no particles are created or destroyed, the atoms are simply rearranged from the reactants to the products.

Oxygen (element) has boiling point of -183°C and hydrogen has boiling point of -253°C. In this chemical change water (compound) is produced and it has new boiling point, boiling point of water is 100°C.

6 0

3 years ago

Read 2 more answers

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