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2 years ago

What is the density of a cube of water measuring 2cmX4cmX1cm, with a mass of 8g?

1 answer:

7 0

The actual formula for volume for a cube is the length multiplied by the width and then multiplied by the height. Since all three measurements are the same, the formula results in the measurement of one side cubed. For the example, 5^3 is 125 cm^3. Multiply the volume by the known density, which is the mass per volume.

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Which base(s) is weaker than ammonia?

Dominik [7]

Answer:

Hydroxylamine and pyridine

Explanation:

Just did it

8 0

3 years ago

How does the chemical formula for the sulfite ion differ from the chemical formula for the sulfate ion?

VLD [36.1K]

Answer:

Sulfite contains 3 oxygens and sulfate contains 4 oxygens

Explanation:

6 0

3 years ago

HCl (hydrochloric acid) is a strong acid; its ions completely dissociate when placed in solution. HF (hydrofluoric acid) is a we

FinnZ [79.3K]

Answer:

The stronger electrolyte is the HCl

Explanation:

Stronger electrolyte are the ones, that in water, completely dissociates.

HCl(aq) → H⁺(aq) + Cl⁻(aq)

HCl(aq) + H₂O(l) → H₃O⁺ (aq) + Cl⁻(aq)

Both are acids, they bring protons to medium but the hydrochloric completely dissociates.

HF (aq) + H₂O(l) ⇄ H₃O⁺(aq) + F⁻(aq) Ka

In the dissociation of weak electrolytes, they ionize but at the same time they bond again, so the reaction is always kept in equilibrium.

5 0

3 years ago

Consider the following reaction of the commercial production of SO3 using sulfur dioxide and oxygen: SO2(g) + O2(g) → SO3(g)

kipiarov [429]

Answer:

$m_{SO_3}=18.93gSO_3$

$V_{SO_3}=5.3LSO_3$

Explanation:

Hello,

STP conditions are P=1 atm and T=273.15 K, thus, the reacting moles are:

$n_{SO_2}=\frac{5.3L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2366molSO_2\\n_{O_2}=\frac{4.7L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2098molO_2$

Now, the balanced chemical reaction turns out into:

$2SO_2(g) + O_2(g) --> 2SO_3(g)$

Thus, the exact moles of oxygen that completely react with 0.2366 moles of sulfur dioxide are (limiting reagent identification):

$0.2366molSO_2*\frac{1molO_2}{2molSO_2}=0.1182molO_2$

Since 0.2098 moles of oxygen are available, we stipulate the oxygen is in excess and the sulfur dioxide is the limiting reagent. In such a way, the yielded grams of sulfur trioxide turn out into:

$m_{SO_3}=0.2366molSO_2*\frac{2molSO_3}{2molSO_2}*\frac{80gSO_3}{1molSO_3} \\m_{SO_3}=18.93gSO_3$

By using the ideal gas equation, one computes the volume as:

$V_{SO_3}=\frac{mRT}{MP}=\frac{18.93g*0.082\frac{atm*L}{mol*K} *273.15K}{80g/mol*1atm}\\V_{SO_3}=5.3LSO_3$

It has sense for volume since the mole ratio is 2/2 between sulfur dioxide and sulfur trioxide.

Best regards.

6 0

3 years ago

Será lançado na próxima quarta-feira, 1º de julho, em São Bernardo do Campo (SP), o primeiro ônibus brasileiro a hidrogênio. [..

slamgirl [31]

Sera bu they are going there for fun

4 0

3 years ago