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3 years ago

Is vinegar a compound, element, or mixture? Thanks a bunch! :))

2 answers:

6 0

Vinegar is a liquid mixture composed of acetic acid and water. It is produced by fermenting ethanol which causes bacteria to produce acetic acid as a byproduct.

xeze [42]3 years ago

5 0

Vinegar is a mixture of acetic acid and water

You might be interested in

Decide which element probably has a boiling point most and least similar to the boiling point of cesium.

natka813 [3]

Answer:

Take a look at the attachment below

Explanation:

Take a look at the periodic table. As you can see, Rubidium is the closest element to Cesium, and happens to have the closest boiling point to Cesium, with only a difference of about 30 degrees.

Respectively, you would think that fluorine should have the least similarity to Cesium with respect to it's boiling point, considering it is the farthest away from the element out of the 4 given. This is not an actual rule, there are no fixed trends of boiling points in the periodic table, there are some but overall the trends vary. However in this case fluorine does have the least similarity to Cesium with respect to it's boiling point, a difference of about 1,546.6 degrees.

Hope that helps!

5 0

3 years ago

Name the method that used to separated amino acids from fruit juice solution

Ket [755]

Using the method chromatography

8 0

3 years ago

CAN SOMEONE PLEASE HELP ME!!!! I CAN'T GET ANY HELP IN THE SCIENCE SECTION!!!!!

lorasvet [3.4K]

Answer: shining a bright light on the object and testing for decomposition

Decomposition is the separation of a molecule into its elements or simpler compounds.

And a chemical property is the ability of a substance to combine or to become one or more substances.

Then, testing the ability of an object for decomposition is testing a chemical property.

6 0

4 years ago

Read 2 more answers

How many molecules are there in 8.00 grams of glucose, C6H12O6

beks73 [17]

Answer:

molecules C6H12O6 = 2.674 E22 molecules.

Explanation:

from periodic table:

⇒ molecular mass C6H12O6 = ((6)(12.011)) + ((12)(1.008)) + ((6)(15.999))

⇒ Mw C6H12O6 = 180.156 g/mol

⇒ mol C6H12O6 = (8.00 g)(mol/180.156 g) = 0.0444 mol C6H12O6

∴ mol ≡ 6.022 E23 molecules

⇒ molecules C6H12O6 = (0.0444 mol)(6.022 E23 molecules/mol)

⇒ molecules C6H12O6 = 2.674 E22 molecules

6 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