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2 years ago

Calcular la normalidad de 1 Kg de sulfuro de aluminio en 5000 ml de solucion.

Chemistry

1 answer:

Troyanec [42]2 years ago

7 0

Calculate the normality of 1 Kg of aluminum sulfide in 5000 ml of solution.

Normality comes out to be 8.11

<h3> Given </h3>

Mass of solute: 1000g
Volume of solution (V): 5000 ml = 5 liters
Equivalent mass of solute (E) = molar mass / n-factor

n-factor for $Al_{2}S_{3}$ is 6 and molar mass is 148g

So, on calculating equivalent mass is equal to 24.66g

FORMULAE of Normality (N) = (Mass of the solute) / (Equivalent mass of the solute (E) × Volume of the solution (V)

N= $\frac{1000}{24.66*5}$

Therefore, normality of 1 kg aluminum sulfide is 8.11

Learn more about normality here brainly.com/question/25507216

#SPJ10

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3 years ago

using the equation you wrote determine how many moles of butane c4h10 are needed to react with 5.5 moles of oxygen

GaryK [48]

Answer:

0.846 moles.

Explanation:

This is a stichiometric problem.

The balanced equation of complete combustion of butane is:

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<u><em>Using cross multiplication:</em></u>

1.0 mole of C₄H₁₀ reacts with → 6.5 moles of O₂

??? moles of C₄H₁₀ are needed to react with → 5.5 moles of O₂

The number of moles of C₄H₁₀ that are needed to react with 5.5 moles of O₂ = (1.0 x 5.5 moles of O₂) / (6.5 moles of O₂) = 0.846 moles.

3 0

3 years ago

A 10-gram aluminum cube absorbs 677 joules when its temperature is increased from 50°C to 125°C. What is the specific heat of al

ra1l [238]

<h3>Answer:</h3>

0.90J/g°C

<h3>Explanation:</h3>

We are given:

Mass of Aluminium = 10 g

Quantity of heat = 677 Joules

Change in temperature = 125°C - 50°C

= 75°C

We are required to calculate the specific heat capacity of Aluminium

But, Quantity of heat = Mass × specific heat × Change in temperature

Q = mcΔt

Rearranging the formula;

c = Q ÷ mΔt

= 677 J ÷ (10 g × 75°C)

= 677 J ÷ 750g°C

= 0.903 J/g°C

= 0.90J/g°C

Thus, the specific heat capacity of Aluminium is 0.90J/g°C

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3 years ago

What are the main properties of liquids (in contrast to gases and solids)? check all that apply. check all that apply. liquids h

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4 years ago

What is the hydronium ion concentration of a 0.100 M acetic acid solution with Ka = 1.8 × 10-5? The equation for the dissociatio

evablogger [386]

Answer:

1.3×10⁻³ M

Explanation:

Hello,

In this case, given the dissociation reaction of acetic acid:

$CH_3CO_2H(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CH_3CO_2^-(aq)$

We can write the law of mass action for it:

$Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}$

Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change $x$ due to the dissociation extent, we are able to rewrite it as shown below:

$1.8x10^{-5}=\frac{x*x}{0.100-x}$

Thus, via the quadratic equation or solve, we obtain the following solutions:

$x_1=-0.00135M\\x_2=0.00133M$

Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.

Regards.

8 0

4 years ago

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