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4 years ago

On what day the lining of the uterus builds up

Chemistry

2 answers:

yan [13]4 years ago

8 0

28 days..................................................................................................

Mariulka [41]4 years ago

7 0

28 days 3 2 2 2 3 3 3 do not mind these answers

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In this reaction, _____.

ella [17]

Answer:

AB + CD ----> AC + BD

Explanation:

If you think this reaction:

AB + CD ----> AC + BD

(Reactants) (Products)

All the statements are true.

7 0

3 years ago

Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO (s) + H2O (l) â†’ Ca(OH)2 (s) In a p

dybincka [34]

Answer:

The option closest to the percentage yield is option;

d. 81.1

Explanation:

The given chemical equation of the reaction is presented as follows;

CaO (s) + H₂O (l) → Ca(OH)₂

The mass of CaO in the experiment, m = 2.00 g

The volume of water with which the CaO was reacted = Excess volume of water

Number of moles = Mass/(Molar mass)

The mass of Ca(OH)₂ recovered, actual yield = 2.14 g

The molar mass of CaO = 56.0774 g/mol

The number of moles of CaO in the reaction, n₁ = 2.00 g/(56.0774 g/mol ≈ 0.036 moles

The molar mass of Ca(OH)₂ = 74.093 g/mol

The number of moles of Ca(OH)₂ in the reaction, n₂ = 2.14 g/(74.093 g/mol) ≈ 0.029 moles

From the given chemical reaction, one mole of CaO reacts with one mole of H₂O to produce one mole of Ca(OH)₂

Therefore, 0.036 moles of CaO will produce 0.036 moles of Ca(OH)₂

Mass = Number of moles × Molar mass

The mass of 0.036 moles of Ca(OH)₂ ≈ 0.036 moles × 74.093 g/mol = 2.667348 grams

∴ The theoretical yield of Ca(OH)₂ = 2.667348 grams

$Percentage \ yield = \dfrac{Actual \ yield}{Theoretical \ yield} \times 100 \%$

The percentage yield = (2.14 g)/(2.667348 grams) × 100 = 80.23%

Therefore, the option which is closest is option d. 81.1.

5 0

3 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

3 years ago

How many moles of fe2o3 will be produced from 18.0 g of fe assuming o2 is available in excess

Advocard [28]

Answer:

0.161moles

Explanation:

Given parameters:

Mass of Fe = 18g

Oxygen gas is in excess

Unknown:

Number of moles of Fe₂O₃ produced = ?

Solution:

To start with, let us write a chemically balanced equation before proceeding to understand the nuances of this problem.

4Fe + 3O₂ → 2Fe₂O₃

In the equation above above, 4 mole of iron combined with 3 moles of oxygen gas to 2 moles of Fe₂O₃.

In solving this problem, we can identify that Fe is the limiting reactant since we have been told oxygen gas is in excess. The suggests that the extent to which the product is formed and the reaction proceeds hinges on the amount of Fe we have.

It is best to work from the given, or known reactant to the unknown

The known in this scenario is the mass of Fe. Let us find the number of moles of this specie;

Number of moles of Fe = $\frac{mass}{molar mass}$

Molar mass of Fe = 56g/mol

Number of moles = $\frac{18}{56}$ = 0.32mol

Using this known number of moles of Fe, we can relate it to that of the unknown amount of the product and obtain the number of moles.

4 moles of Fe produced 2 moles of Fe₂O₃

0.32 moles of Fe will produce $\frac{0.32 x 2}{4}$ = 0.161moles

8 0

3 years ago

Which of the following is not true about compounds?

finlep [7]

Answer:

Elements can combine in any proportion to form a compound. -third choice

4 0

3 years ago

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