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2 years ago

4. What is the difference between an open system and a closed system?

1 answer:

5 0

Systems can exist in three ways as open systems, closed systems, and isolated systems. The main difference between open and closed system is that in an open system, matter can be exchanged with the surrounding whereas, in a closed system, matter cannot be exchanged with the surrounding.

btw why not just look it up on google? btw btw brainliest plz

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Which of these pairs of elements have the same number of valence electrons?

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The correct answer is the second option; sodium (Na) and potassium (K.)
Both sodium and potassium have the same number of valence electrons.

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2 years ago

Which set of elements is correctly matched with its ionic formula?

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2 years ago

Instructions:Select the correct answer.

andriy [413]

Activity series of metals: K,Na,Mg,Al,Zn,Fe,Cu,Ag. Metals on the left are more reactive than metals on the right. For example Zn is more reactive than Fe and can displace him.
Reaction than can occur is: <span>CuSO4(aq) + Fe(s) → FeSO4(aq) + Cu(s).</span>

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2 years ago

Light energy is a form of A. neither potential nor kinetic energy. B. kinetic energy only. C. both potential and kinetic energy.

GarryVolchara [31]

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2 years ago

Read 2 more answers

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

2 years ago