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2 years ago

Write out the recipe and list all the ingredients (including the measures) of your favorite dish.

Chemistry

1 answer:

arsen [322]2 years ago

4 0

The recipe for an omelet are:

Ingredients

1 Egg
Salt and pepper ( 1 tablespoon )
Butter ( 1 tablespoon )

Recipe:

Turn on the heat and place a pan on it.
Add butter to the pan.
Crack the egg and whisk it properly.
Add the whisked omelet onto the pan and stir.
Add salt and pepper.

<h3>How to illustrate the information?</h3>

When we turn on the heat and place the pan, a heat transfer takes place and heat transfers from one body to another.

When we add butter to the pan, then the temperature of the pan was more than that of the butter.

Since the melting point of butter was less than the temperature of the pan, the the melting of butter takes place.

When we add the egg, the evaporation of the liquid present in the egg yolk and the egg white takes place.

These are the chemical reactions depicted.

Learn more about recipe on:

brainly.com/question/21913357

#SPJ1

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WILL GIVE 50 POINTS PLEASE HELP!!!!!

Brut [27]

Answer:

b) The volume of stock solution used is less than 3 liters.

Explanation:

Given:

Concentration of stock solution of NaCl, M1 = 10 M

Concentration of diluted solution of NaCl, M2 = 1 M

Volume of diluted NaCl solution, V2 = 3 L

Let the volume of stock NaCl solution = V1

Formula:

Use the dilution relation:

$M1V1 = M2V2$

Based on the above equation:

$V1 = \frac{M2V2}{M1} = \frac{1M*3L}{10M} = 0.3 L$

Therefore, the volume of stock solution required is 0.3 L which is less than 3 L

6 0

3 years ago

Considering the following precipitation reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq),

patriot [66]

Answer: The ions which are not spectator ions are $Pb^{2+}$ and $I^-$

Explanation:

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

The given chemical equation is:

$Pb^{2+}(aq)+NO_3^-(aq)+2K^+(aq)+2I^-(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^-(aq)$

The ions which are present on both the sides of the equation are $K^+$ and $NO_3^-$ and are spectator ions.

Thus the net ionic equation is:

$Pb^{2+}(aq)+2I^-(aq)\rightarrow PbI_2(s)$

Hence, the correct answer is $Pb^{2+}$ and $I^-$

5 0

3 years ago

Read 2 more answers

Condensation point and freezing point of argon in KELVIN.

vesna_86 [32]

Answer:

Condensation: 423.3 K

Freezing: 83.96 K

(this is all i could figure out :) hope it helps)

4 0

3 years ago

Zinc metal is added to a flask containing aqueous hydrochloric acid. The flask contains 0.400 mole of HCl. How much hydrogen gas

dusya [7]

Answer:

0.4 g of hydrogen gas would be produced.

1. 0.48 mole of HCl is needed to react completely with 15.5 g of zinc.

2. HCl is the limiting reactant.

3. 2.61 g of zinc is in excess

Explanation:

From the balanced equation of reaction:

$Zn(s)+ 2HCl(aq) --> H_2(g) + ZnCl_2(aq)$

1 mole of Zn requires 2 moles of HCl to produce 1 mole of hydrogen gas.

15.5 g of zinc = 15.5/65.3 = 0.24 moles of zinc.

0.24 moles Zn is supposed to require 0.24 x 2 moles HCl which is equivalent to 0.48 moles HCl.

But only 0.400 mole of HCl is present. Hence, HCl is the limiting reagent. For complete reaction with 15.5 g of Zinc, 0.48 mole HCl would be needed.

0.400 mole of HCl will require 0.2 mole of Zn for complete reaction. This thus means that 0.24 - 02 = 0.04 mole of Zn is in excess.

0.04 mole Zn = 0.04 x 65.3 = 2.61 g excess Zn.

Now, since HCl is the limiting reagent;

2 moles of HCl is required to produce 1 mole of H2 according to the equation.

0.400 mole HCl will therefore yield 0.400 x 1/2 = 0.2 mole H2

0.2 mole H2 = 2 x 0.2 = 0.4 g H2

Hence, 0.4 g of hydrogen gas would be produced.

6 0

3 years ago

Calculate the cell potential for the reaction as written at 25.00 °C 25.00 °C , given that

Taya2010 [7]

Answer:

E = 2.02 V

Explanation:

In order to do this, we need to apply the Nernst equation which is:

E = E° - RT/nF lnQ

The value of RT/F can be simplified to just 0.059 because we are doing this experiment at 25 °C, and R and F are constants. so we need the value of Q which in this case is:

Q = [Mg²⁺] / [Ni²⁺]

We already have the concentrations, so, all we have left is the standard reduction potential, which are:

E° Mg = -2.38 V

E° Ni = -0.25 V

According to the overall reaction:

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)

we can see that one element is reducting and the other is oxidizing, so we need to write the semi equation of reduction for each element:

Mg(s) ---------> Mg²⁺ + 2e⁻ E° = 2.38 V oxidizing (Value of E° inverted)

Ni²⁺ + 2e⁻ -----------> Ni(s) E° = -0.25 V reducting

------------------------------------------------------------

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s) E° = 2.13 V

We have the value of the standard potential, now we need to replace all given data into the nernst equation to solve for the cell potential:

E = 2.13 - 0.059/2 ln(0.757/0.0160)

E = 2.13 - 0.0295 ln(47.3125)

E = 2.13 - 0.11

E = 2.02 V

This is the cell potential

3 0

3 years ago