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3 years ago

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When reacting 45g of magnesium with oxygen, 50.0g of magnesium oxide is produced. Calculate the theoretical yield and percent yi

eld.

1 answer:

Veseljchak [2.6K]3 years ago

5 0

Hope you are able to get it :)

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At the start of a reaction, there are 0.0249 mol N2,

gladu [14]

Answer:

Explanation:

The reaction is given as:

$N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$

The reaction quotient is:

$Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}$

From the given information:

TO find each entity in the reaction quotient, we have:

$[NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}$

$[N_2] = \dfrac{0.024 }{3.5}$

$[N_2] = 0.006857$

$[H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}$

$[H_2] = 9.17 \times 10^{-3}$

∴

$Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135$

However; given that:

$K_c = 1.2$

By relating $Q_c \ \ and \ \ K_c$ , we will realize that $Q_c \ \ < \ \ K_c$

The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.

7 0

3 years ago

Which member of each pair is more soluble in diethyl ether? Why?<br> (c) MgBr₂(s) or CH₃CH₂MgBr(s)

White raven [17]

CH3CH2MgBr is more soluble in diethyl ether .

We know that polar solvent dissolve in polar solvent very perfectly . as diethyl ether is a polar solvent so it have dipole -dipole interaction .

Hence the compound with similar interaction can dissolve in diethyl ether .

Here , MgBr2 is an ionic compound . there is ion-ion interactions occurs which is not similar to dipole -dipole interaction in diethyl ether .hence the solubility of MgBr2 in diethyl ether is less .

but in case of CH3CH2MgBr there are both polar and nonpolar end .CH3CH2 is the nonpolar end and MgBr is the polar end .

thus with the nonpolar end solute interact using depression forces and with polar end solute interact using dipole-dipole interaction . so CH3CH2MgBr is more soluble .

Learn more about polar solvent here :

brainly.com/question/3184550

#SPJ4

4 0

2 years ago

6.0 moles of Na and 4.0 moles of Cl2 are mixed, how manu moles of NaCl in moles cane be made from this mixture

Sladkaya [172]

Answer:

Moles of NaCl formed is 6.0 moles

Explanation:

We are given the equation;

2 Na(s) + Cl₂(g) → 2 NaCl(s)

Moles of Na is 6.0 moles
Moles of Cl₂ is 4.0 moles

From the reaction;

2 moles of sodium reacts with 1 mole of chlorine gas to form 2 moles of NaCl

In this case;

6 moles of Na would require 3 moles of Cl₂, this means that chlorine gas is in excess.

Thus, the rate limiting reagent is sodium.

But, 2 moles of sodium reacts to form 2 moles of NaCl

Therefore;

Moles of NaCl = Moles of Na

= 6.0 moles

Thus, moles of sodium chloride produced is 6.0 moles

3 0

3 years ago

how many moles of iodine should be added to 750 grams of carbon tetrachloride to prepare a 0.24 m solution

aleksley [76]

Answer:

0.18 mol

Explanation:

Given data

Mass of carbon tetrachloride (solvent): 750 g

Molality of the solution: 0.24 m

Moles of iodine (solute): ?

Step 1: Convert the mass of the solvent to kilograms

We will use the relationship 1 kg = 1,000 g.

$750g \times \frac{1kg}{1,000g} =0.750kg$

Step 2: Calculate the moles of the solute

The molality is equal to the moles of solute divided by the kilograms of solvent. Then,

$m = \frac{moles\ of\ solute }{kilograms\ of\ solvent} \\moles\ of\ solute = m \times kilograms\ of\ solvent = \frac{0.24mol}{kg} \times 0.750kg = 0.18 mol$

6 0

3 years ago

Read 2 more answers

Match each transition metal ion with its condensed ground-state electron configuration. A [Ar]3d2 B [Ar]4s23d3 C [Kr]4d10 D [Xe]

madreJ [45]

Answer:

$Mn^{2+}$ : - F . $[Ar]3d^{5}$

$Hg^{2+}$ : - G. $[Xe]4f^{14}5d^{10}$

$La^{3+}$ : - D. $[Xe]$

$Fe^{3+}$ : - F. $[Ar]3d^{5}$

$Ag^{+}$ : - C. $[Kr]4d^{10}$

$Co^{3+}$ : - E. $[Ar]3d^{6}$

Explanation:

The electronic configuration of the element Mn is:-

$[Ar]3d^{5}4s^2$

For, $Mn^{2+}$ , 2 electrons are lost, thus the configuration is:-

$[Ar]3d^{5}$

The electronic configuration of the element Hg is:-

$[Xe]4f^{14}5d^{10}6s^2$

For, $Hg^{2+}$ , 2 electrons are lost, thus the configuration is:-

$[Xe]4f^{14}5d^{10}$

The electronic configuration of the element La is:-

$[Xe]5d^{1}6s^2$

For, $La^{3+}$ , 3 electrons are lost, thus the configuration is:-

$[Xe]$

The electronic configuration of the element Fe is:-

$[Ar]3d^{6}4s^2$

For, $Fe^{3+}$ , 3 electrons are lost, thus the configuration is:-

$[Ar]3d^{5}$

The electronic configuration of the element Ag is:-

$[Kr]4d^{10}5s^1$

For, $Ag^{+}$ , 1 electron is lost, thus the configuration is:-

$[Kr]4d^{10}$

The electronic configuration of the element Co is:-

$[Ar]3d^{7}4s^2$

For, $Co^{3+}$ , 3 electrons are lost, thus the configuration is:-

$[Ar]3d^{6}$

7 0

3 years ago