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3 years ago

Please Help!!

Chemistry

2 answers:

Dmitry [639]3 years ago

8 0

Answer:

weak

Explanation:

Leokris [45]3 years ago

6 0

Answer: The covalent bonds within a molecule are very (strong). However it requires (less) energy to melt a molecular compound than an ionic compound.

This is because the attractive forces between individual molecules are (weaker) than the forces between ions in an ionic compound.

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If 23.7 g of Al(OH)3(s) are mixed with 29.5 g of H2SO4(s) and the reaction is run, answer the following questions:

expeople1 [14]

Aluminum hydroxide $\text{Al}(\text{OH})_3$ can behave as a base and neutralize sulfuric acid $\text{H}_2\text{SO}_4$ as in the following equation:

$2\;\text{Al}(\text{OH})_3 \; (s) + 3\; \text{H}_2\text{SO}_4 \; (aq) \to \text{Al}_2(\text{SO}_4)_3 \; (aq) + 6 \; \text{H}_2\text{O} \; (l)$ (Balanced)

(a)

$n = m/M$ . Thus the ratio between the number of moles of the two reactants available:

$n(\text{Al}(\text{OH})_3, \text{supplied}) / n(\text{H}_2\text{SO}_4, \text{supplied})\\= [m(\text{Al}(\text{OH})_3)/ M(\text{Al}(\text{OH})_3)] / [n(\text{H}_2\text{SO}_4) / M(\text{H}_2\text{SO}_4)]\\= [23.7 / (26.98 + 3 \times(16.00 + 1.008))]/[29.5 / (2 \times 1.008 + 32.07 + 4 \times 16.00)]\\\approx 1.01$

The value of this ratio required to lead to a complete reaction is derived from coefficients found in the balanced equation:

$n(\text{Al}(\text{OH})_3, \text{theoretical}) / n(\text{H}_2\text{SO}_4, \text{theoretical}) = 2/3 \approx 0.667$

The ratio for the complete reaction is smaller than that of the reactants available, indicating that the species represented on the numerator, $\text{Al}(\text{OH})_3$ , is in excess while the one on the denominator, $\text{H}_2\text{SO}_4$ , serves as the limiting reagent.

(b)

The quantity of water produced is dependent on the amount of limiting reactants available. $29.5 / (2 \times 1.008 + 32.07 + 4 \times 16.00) = 0.301 \; \text{mol}$ of sulfuric acid is supplied in this reaction as the limiting reagent. $6$ moles of water molecules are produced for every $3$ moles of sulfuric acid consumed. The reaction would thus give rise to $0.301 \; \text{mol} \times 6/3 = 0.602 \; \text{mol}$ of water molecules, which have a mass of $0.602 \times (2 \times 1.008 + 16.00) = 10.8 \; \text{g}$ .

(c)

$\text{Percentage Yield}\\= \text{Actual Yield} / \text{Theoretical Yield} \times 100 \; \%\\= 2.21 / 10.8 \times 100 \; \%\\= 20.4 \; \%$

(d)

The quantity of $\text{Al}(\text{OH})_3$ , the reactant in excess, is dependent on the number of moles of this species consumed in the reaction and thus the quantity of the limiting reagent available. The consumption of every $3$ moles of sulfuric acid, the limiting reagent, removes $2$ moles of aluminum hydroxide $\text{Al}(\text{OH})_3$ from the solution. $0.301 \; \text{mol}$ of sulfuric acid is initially available as previously stated such that $0.301 \; \text{mol} \times 2/3 = 0.201 \; \text{mol}$ , or $0.201 \times (26.98 + 3 \time (16.00 + 1.008)) = 15.7 \; \text{g}$ , of $\text{Al}(\text{OH})_3$ would be eventually consumed.

$23.7 - 15.7 = 8.0 \; \text{g}$ of $\text{Al}(\text{OH})_3$ would thus be in excess by the end of the reaction process.

3 0

3 years ago

How many moles of KNO3 are needed to make 600 ml of a 1.3M solution?

ludmilkaskok [199]

M = n / V

Where, M is molarity (M or mol/L), n is number of moles of the solute (mol) and V is volume of the solution (L).

Here the solute is KNO₃.
The given molarity is 1.3 M
This means 1L of solution has 1.3 moles of KNO₃.

Hence moles in 600 mL = 1.3 M x 0.6 L = 0.78 mol

Therefore to make 1.3 M KNO₃ solution, needed moles of KNO₃ is 0.78 mol

7 0

3 years ago

Ms. Salis explains to her class that sodium (Na) and chlorine (Cl) atoms bond to make table salt

larisa [96]

The answer is D a compound :)

6 0

3 years ago

n-Butane (C4H10) is burned with stoichiometric amount of oxygen. Determine the mole fraction of carbon dioxide and water in the

Fudgin [204]

Answer:

See details below

Explanation:

The balanced reaction equation is given below:

$2C_{4} H_{10}$ + $13O_{2}$ → $8CO_{2}$ + $10H_{2} O$

Mole fraction of CO2 to H20

= 8/10 = $\frac{4}{5}$

Mole ratio of C4H10 to CO2 is 2:8 = 1:4

1 mole of n-butane - 38.12 g

4 moles - ?

= 152.48g fuel consumed.

8 0

3 years ago

** ANSWER FAST I WILL MARK YOU BRAINLIEST AND GIVE 20 POINTS * * THIS IS A CHEMISTRY QUESTION SO PLEASE ONLY ANSWER IF YOU ARE G

Anna11 [10]

Answer:

$\boxed{\text{ B. Increase the temperature and decrease the pressure.}}$

Explanation:

Let's say the reaction is

R ⇌ 2P; endothermic

I like to consider heat as if it were a reactant or a product in a chemical equilibrium.

Another way to write the equilibrium would be

heat + R ⇌ 2P

According to Le Châtelier's Principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Let's consider each of the stresses in turn.

(i) Changing the temperature

If you want to increase the amount of product, you increase the temperature. The system will try to get rid of the added heat by shifting to the right, thus forming more product.

(ii) Changing the pressure

If R and P are liquids or solids or in aqueous solution, changing the pressure will have no effect. Something must be in the gas phase for a change in pressure to affect the position of equilibrium.

If P is a gas, the equilibrium is

heat + R ⇌ 2P(g)

Then, decreasing the pressure will produce more P. If you reduce the pressure, the system will respond by shifting to the right (the side with more gas molecules) to produce more P and bring the pressure back up

5 0

3 years ago