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3 years ago

I need this fast plz, 50 points

Chemistry

2 answers:

attashe74 [19]3 years ago

6 0

Answer:

It is decreasing the temperature of the reactant mixture.

Explanation:

Just took the test

Margarita [4]3 years ago

4 0

Answer:

increasing the pressure of the reactant mixture

Explanation:

according to the collision theory , the number of collisions that take place per unit volume of the reaction mixture is called collision frequency. The effective collisions are ones which result into the formation of products. It depends on energy factor: For collision to be effective, the colliding molecules must have energy more than a particular value called as threshold energy. increasing the temperature of the reactant mixture, the reactant molecules will gain kinetic energy and thus will increase the number of effective collisions.

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Which of these elements have the strongest and the weakest forces between its atoms.Give A Reason For Your Answer. Copper,oxygen

fiasKO [112]

Answer:

Copper

Explanation:

Within intermolecular forces, ion-dipole is the strongest, followed by hydrogen bonding, then dipole-dipole, and then London dispersion.

4 0

2 years ago

Identify the precipitate that forms (if any) when aqueous solutions of barium nitrate and sodium sulfate are mixed.

Alborosie

The precipitate that forms when aqueous solutions of barium nitrate and sodium sulfate are mixed is barium sulphate.

Barium Nitrate is a colorless to white, odorless, crystalline powder. It is used in making fireworks, for green lights and neon lights, and in ceramic glazes.

Sodium Sulfate Anhydrous is the anhydrous, sodium salt form of sulfuric acid. Sodium ion is the principal cation of the extracellular fluid and plays a large part in the therapy of fluid and electrolyte disturbances.

When aqueous solutions of barium nitrate and sodium sulfate are mixed is barium sulphate, one soluble salt, sodium nitrate, and an insoluble salt, barium sulphate.

Learn more about barium nitrate, click here brainly.com/question/9597126

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4 0

2 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

HELP ME PLEASE!!!! I NEEDZ HELP!!!

Anuta_ua [19.1K]

It goes Meter mega meter giga meter and kilometer

3 0

3 years ago

PLEASE SOMEBODY EXPLAIN THIS :(((

djverab [1.8K]

Answer:

$\boxed{ \sf \: R_f \: value \: of \: sample \: 1 =0.3142}$

<h3>Explanation:</h3>

In Analytical Chemistry chromatography is widely used for the separation of samples.

In thin layer chromatography, the mixture of components are separated on the basis of their polarity.
The solvent solution(mobile phase) that we use are non polar & silica gel( TLC paper made of/stationary phase) are polar.
Consider the mixture we have taken consist of two samples having large polar difference.
Due to opposite nature of silica gel(polar) & solvent solution (non polar) the movement become easy & due to capillary action solvent solution rise to the top.
The mixture of sample we have taken, the sample have less polarity have high peak or they travel more distance than that of more polar sample when they dipped into the solution.

In the given diagram, mixture of 8 samples are separated on the basis of their polarity, the distance travelled by solvent is 35 mm, distance travelled by sample 1 is 11 mm & similarly distance travelled by sample 2,3,4,5,6,7 are 15,31,4,22,25,33 in mm respectively.

Rf Value: Rf value is retention factor which tells about relative absorption of each sample & range of Rf value is 0-1.

Formula to calculate Rf value is

$\sf R_f \: value = \frac{distance \: moved \: by \: sample}{distance \: moved \: by \: solvent}$

Now, solving for Rf value of sample 1

Given:

Distance moved by sample 1 = 11 mm

Distance movedby solvent = 35 mm

To find:

Rf value of sample 1 = ?

Solution:

Substituting the given data in above formula,

$\small \sf R_f \: value = \frac{distance \: moved \: by \: sample \: 1}{distance \: moved \: by \: solvent} \\ \small \sf R_f \: value = \cancel\frac{11 \: mm}{35 \: mm} = 0.3142$

$\small \boxed{ \sf \: R_f \: value \: of \: sample \: 1 =0.3142}$

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4 0

2 years ago