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1 year ago

Question 3 options: how many moles of argon atoms are present in 11.2 l of argon gas at stp?

1 answer:

5 0

It is known that at if a gas is identified to be as ideal gas then, at STP, the volume of 1 mole of that gas is equal to 22.4 L. Using this fact and the given volume in this item, we determine the number of moles as that below.

n = (11.2 L of argon)(1 mol argon / 22.4 L of argon)

n = 0.5 moles

Answer: 0.5 moles

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An experiment requires that each student use an 8.5 cm length of magnesium ribbon. How many students can do the experiment if th

melisa1 [442]

570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students

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1 year ago

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. Calculate the mass of O2 produced if 3.450 g potassium chlorate is completely decomposed by heating in presence of a catalyst

Vesnalui [34]

2 KClO3(s) → 3 O2(g) + 2 KCl(s)

Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.

molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028

Moles of O2 produce = $\frac{3}{2} \times 0.028$

= 0.042 moles

molar mass of O2 = 32

so, mass of O2 = 32 x 0.042 = 1.35 g

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1 year ago

Use the following half-reactions to design a voltaic cell: Sn4+(aq) + 2 e− → Sn2+(aq) Eo = 0.15 V Ag+(aq) + e−→ Ag(s) Eo = 0.80

AVprozaik [17]

Answer:

E° = 0.65 V

Explanation:

Let's consider the following reductions and their respective standard reduction potentials.

Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V

Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:

Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V

The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V

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1 year ago

Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.

Kay [80]

Answer:

-) 3-bromoprop-1-ene

-) 2-bromoprop-1-ene

-) 1-bromoprop-1-ene

-) bromocyclopropane

Explanation:

In this question, we can start with the I.D.H (hydrogen deficiency index):

$I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}$

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

$I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1$

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (1-bromoprop-1-ene). In the second structure, we can move the Br atom to the second carbon (2-bromoprop-1-ene), in the third structure we can move the Br to carbon 3 (3-bromoprop-1-ene). Finally, we can have a cyclic structure with a Br atom (bromocyclopropane).

See figure 1

I hope it helps!

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1 year ago

How do models help scientists predict the polarity of molecules?

yuradex [85]

Answer:There are three main properties of chemical bonds that must be considered—namely, their strength, length, and polarity. The polarity of a bond is the distribution of electrical charge over the atoms joined by the bond. Specifically, it is found that, while bonds between identical atoms (as in H2) are electrically uniform in the sense that both hydrogen atoms are electrically neutral, bonds between atoms of different elements are electrically inequivalent. In hydrogen chloride, for example, the hydrogen atom is slightly positively charged whereas the chlorine atom is slightly negatively charged. The slight electrical charges on dissimilar atoms are called partial charges, and the presence of partial charges signifies the occurrence of a polar bond.

Explanation:

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1 year ago

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