Ask question

Ask question

All categories

3 years ago

14

Write a procedure for "how to crack an egg" in cooking. You must have at least 3 steps. The format of your procedure is more imp

ortant than the actual steps on how to do it.

1 answer:

Nataly_w [17]3 years ago

6 0

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

Here's the procedure to Crack an egg ~

Step 1 ~

Hold the egg in your hand and tap it on a firm surface

Step 2 ~

Gently pull the shells apart

Step 3 ~

And release the yolk in the bowl

You might be interested in

What type of hybridization is needed to explain why ethyne, C2H2, is linear?

Dmitrij [34]

When C-C is having a triple bond the hybridization is sp. But I am not sure how to relate that to the linear shape.

5 0

4 years ago

Why don't we use meters or miles when discussion distances between stars or galaxies?

taurus [48]

If we use mile or something else it will be hard to measure in sky .
Because the mile or meters too short for the space,
Galaxy and stars are too long far away from us if we use mile or meter,
It till be take like miliion year to measure.

4 0

3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

4 years ago

2. What is the mass of 16.3 L of helium gas?

lilavasa [31]

(16.3 L) / (22.414 L/mol) x (4.0026 g He/mol) = 2.91 g

3 0

3 years ago

Read 2 more answers

In the presence of a base, blue litmus paper will .<br>O turn purple<br>stay blue<br>to turn red

Inessa05 [86]

Answer:

In the presence of a base, blue litmus paper will turn red........

7 0

3 years ago

Read 2 more answers