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Ratling [72]
2 years ago
10

1. How much 6.0 M HNO3 is needed to neutralize 39 mL of 2.0 M KOH?

Chemistry
1 answer:
quester [9]2 years ago
6 0

Answer:

For 2. the answer is 15.0 mL

For other examples, you can solve by exact way as I have solved the 2nd example.

I have writen down all the balanced chemical reaction equation for examples 1, 3, 4, 5 for you. ( picture 2 )

Explanation:

Please see the step-by-step solution in the picture attached below.(picture 1)

Hope this answer can help you. Have a nice day!

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Out of magnesium and Calcium which element is more reactive give reason​
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Answer:

Calcium is more reactive.

Explanation:

It is higher in the reactivity series, and contains more energy levels.

5 0
2 years ago
As a real gas deviates from ideal gas behavior, the particles _____
Alik [6]

Answer:

Have some attraction towards each other

Explanation:

Gases deviate from the ideal gas behavior because their molecules have forces of attraction between them. At high pressure, the molecules of gases are very close to each other so the molecular interactions start operating and these molecules do not strike the walls of the container with full impact.

Hope this helps :-)

Have a great rest of your day or night!

Enjoy your studies and assignments

<3 simplysun

ps. I do not own any of these answers so please don't give full credit to me

7 0
2 years ago
Why are the Flemish called Flemish
Dmitriy789 [7]

Explanation:

The flemish are an ethnic group which belong to the northern part of Belgium called Flanders. They also speak a language called flemish. The major population of Belgium are the flemish. The language they speak is a  Low Franconian dialect cluster of the Dutch language.

There was a country called the flanders located in the lowlands of Europe. Any person from this country was considered as flemish previously.

4 0
2 years ago
Ex of physical Properties
sesenic [268]

Answer:

Explanation:

physical properties include:

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etc............

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
3 years ago
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