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hodyreva [135]
4 years ago
7

A vegetable soup recipe requires one teaspoonful of salt. A chef accidentally puts in one tablespoonful. Now the soup is much to

o salty.
a) What can the chef do to reduce the salty taste of the soup?
b) What effects would your suggestion in a) have on the soup?
Chemistry
1 answer:
nirvana33 [79]4 years ago
5 0

Answer:

a. Put a piece of fresh sliced yam with a bore into it into the soup.

Explanation:

b. Osmosis may occur

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Plz I need help is the 2nd time I post this
pychu [463]

Answer:

2--->C

6---->E

3---->D

4--->A

5--->B

1---->F

Explanation:

I think so, sorry if its wrong.

8 0
3 years ago
A sugar crystal contains approximately 1. 9×1017 sucrose (c12h22o11c12h22o11) molecules. part a what is its mass in mgmg ? expre
yKpoI14uk [10]

Sucrose is a disaccharide, which is a sugar molecule formed from two monosaccharides. In this case, the two monosaccharides are fructose and glucose.

Answer and Explanation: 

The number of molecules a" role="presentation" style="display: inline-block; line-height: 0; text-indent: 0px; text-align: left; text-transform: none; font-style: normal; font-weight: 400; font-size: 16.66px; letter-spacing: normal; overflow-wrap: normal; word-spacing: 0px; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; color: rgb(0, 0, 0); font-family: "Open Sans", "Helvetica Neue", Helvetica, Arial, sans-serif; font-variant-ligatures: normal; font-variant-caps: normal; orphans: 2; widows: 2; -webkit-text-stroke-width: 0px; text-decoration-thickness: initial; text-decoration-style: initial; text-decoration-color: initial; position: relative;">aa is related to a quantity of moles n" role="presentation" style="display: inline-block; line-height: 0; text-indent: 0px; text-align: left; text-transform: none; font-style: normal; font-weight: 400; font-size: 16.66px; letter-spacing: normal; overflow-wrap: normal; word-spacing: 0px; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; color: rgb(0, 0, 0); font-family: "Open Sans", "Helvetica Neue", Helvetica, Arial, sans-serif; font-variant-ligatures: normal; font-variant-caps: normal; orphans: 2; widows: 2; -webkit-text-stroke-width: 0px; text-decoration-thickness: initial; text-decoration-style: initial; text-decoration-color: initial; position: relative;">nn of any substance by Avogadro's number {eq}N_A = \rm 6.022\times...

4 0
2 years ago
List down 3 Filipino world-class instrumentalists.<br>1.<br>2.<br>3.​
skad [1K]

Answer:

1)Joy Ayala

2)Gary Granada

3)Levi Celerio

Explanation:

Sana Makatulong

6 0
3 years ago
In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined
faltersainse [42]

<u>Answer:</u>

<u>For A:</u> The standard cell potential of the reaction is 4.4 V

<u>For B:</u> The standard Gibbs free energy of the reaction is -8.50\times 10^5J

<u>For C:</u> The reaction is spontaneous as written.

<u>Explanation:</u>

  • <u>For A:</u>

The given chemical reaction follows:

2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)

The given half reaction follows:

<u>Oxidation half reaction:</u>  Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V ( × 2)

<u>Reduction half reaction:</u>  Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=1.36-(-3.04)=4.4V

Hence, the standard cell potential of the reaction is 4.4 V

  • <u>For B:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

n = number of electrons transferred = 2mol\text{ e}^-

F = Faradays constant = 96500J/V.mol\text{ e}^-

E^o_{cell} = standard cell potential = 4.4 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J

Hence, the standard Gibbs free energy of the reaction is -8.50\times 10^5J

  • <u>For C:</u>

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

7 0
4 years ago
Sodium chloride, NaCl forms in this reaction between sodium and chlorine. 2Na(s) + Cl2(g) → 2NaCl(s) How many moles of NaCl resu
choli [55]

Answer: 7.8 moles of NaCl result from the complete reaction of 3.9 mol of Cl_2

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

\text{Moles of} Cl_2=3.9mol

2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)

As Na is the excess reagent, Cl_2 is the limiting reagent as it limits the formation of product.

According to stoichiometry :

1 mole of Cl_2 gives = 2 moles of NaCl

Thus 3.9 moles of Cl_2 will give=\frac{2}1}\times 3.9=7.8moles  of NaCl

7.8 moles of NaCl result from the complete reaction of 3.9 mol of Cl_2

5 0
3 years ago
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