Ask question

Ask question

All categories

2 years ago

6

Which of the following are elements, which are molecules but not compounds, which are compounds but not molecules, and which are

both compounds and molecules? (a) SO2, (b) S8, (c) Cs, (d) N2O5, (e) O, (f) O2, (g) O3, (h)CH4, (i) KBr, (j) S, (k) P4, (l) LiF.

2 answers:

Pavlova-9 [17]2 years ago

6 0

Answer:

Elements: Cs, O and S.

Molecules: S8, O2, O3 and P4.

Compounds: SO2, N2O5, CH4, KBr and, LiF.

Explanation:

Hello,

At first, remember that elements are present by themselves, molecules are composed by more than one equal element and compounds by more than two different elements. In such a way:

Elements: Cs, O and S.

Molecules: S8, O2, O3 and P4.

Compounds: SO2, N2O5, CH4, KBr and, LiF.

Best regards.

andreev551 [17]2 years ago

4 0

Explanation:

Molecules of compounds are are compounds composed two or more than two different elements.

Molecules of an elements are made up two or more than two numbers of same type of atoms

(a) $SO_2$

It is compound composed of sulfur and oxygen element.

(b) $S_8$

It is a molecules of a single element made up of 8 sulfur atoms.

(c) Cs

It is a symbol of an element named cesium.

(d) $N_2O_5$

It is compound composed of nitrogen and oxygen element.

(e) O

It is a symbol of an element named oxygen.

(f) $O_2$

It is a molecule of a single element made up of 2 oxygen atoms.

(g) $O_3$

It is a molecule of a single element made up of 3 oxygen atoms.

(h) $CH_4$

It is compound composed of carbon and hydrogen element.

(i) KBr

It is compound composed of potassium and bromine element.

(j) S

It is a symbol of an element named sulfur.

(k) $P_4$

It is a molecule of a single element made up of 4 phosphorus atoms.

(l) LiF

It is compound composed of lithium and fluorine element.

You might be interested in

To save time you can approximate the initial volume of water to ±1 mL and the initial mass of the solid to ±1 g. For example, if

WARRIOR [948]

Answer:

20.2 g silver placed in 21.6 mL of water and 20.2 g copper placed in 21.6 mL of water

Explanation:

Since the density formula is mass divided volume, you can calculate the density of each of the components of the group and see which of all the groups are equal

In the group

density Ag=20.2 g silver/21.6ml H2O=0.9352g/ml

density Cu=20.2 g copper/21.6ml H2O=0.9352 g/ml

6 0

2 years ago

Read 2 more answers

If the actual yield of the reaction was 75% instead of 100%, how many molecules of no would be present after the reaction was ov

artcher [175]

To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:

N₂ + O₂ → 2 NO

Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.

1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present

Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>

8 0

2 years ago

Read 2 more answers

For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)?You do not need to look up any values to answer t

Nutka1998 [239]

Answer:

Reactions 1, 3 and 5

Explanation:

First thing's first, let's ensure that all the reactions given are balanced. This is given as;

CO(g) + 1/2 O2(g )→ CO2(g)

Li(s) + 1/2 F2(l) → LiF(s)

C(s) + O2(g) → CO2(g)

CaCO3(g) → CaO + CO2(g)

2Li(s) + F2(g) → 2LiF(s)

For the condition to be valid;

- There is by convention 1 mol of product made. This means we eliminate reactions with more than one mole of compound formed. This eliminates reaction 5.

- The lements haveto be in their state at room temperature. Fluorine is a gas, not a liquid, at room temperature ans pressure, so 2 is not a correct answer.

This leaves us with reactions 1, 3 and 5 as the correct reactions that satisify the condition.

3 0

2 years ago

If 3.491 grams of the precipitate was formed, how many moles of strontium bromide were reacted

AVprozaik [17]

The balanced chemical equation that represents the reaction is as follows:
<span>SrBr2(aq) + 2AgNO3(aq) → Sr(NO3)2(aq) + 2AgBr(s)
</span>
From the periodic table:
mass of silver = 108 grams
mass of bromine = 80 grams

molar mass of silver bromide = 108 + 80 = 188 grams

number of moles = mass / molar mass
number of moles of produced precipitate = 3.491/188 = 0.018 moles

From the balanced equation:
1 mole of strontium bromide produces 2 moles of silver bromide. Therefore, to calculate the number of moles of <span>strontium bromide that produces 0.018 moles of silver bromide, you will just do a cross multiplication as follows:
amount of </span><span>strontium bromide = (0.018x1) / 2 = 9.28 x 10^-3 moles</span>

4 0

2 years ago

A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t

TEA [102]

Explanation:

The reaction equation will be as follows.

$Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)$

So, according to this equation, 1 mole $Ca(OH)_{2}$ = 2 mol HBr = 1 mol $CaBr_{2}$

Therefore, calculate the number of moles of calcium hydroxide as follows.

No. of moles of $Ca(OH)_{2}$ = $V \times Molarity$

= $50 \times 0.6$

= 30 mmol

Similarly, calculate the number of moles of HBr as follows.

No. of moles of HBr = $M \times V$

= $50 \times 0.6$

= 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of $CaBr_{2}$ = $30 \times \frac{1}{2}$

= 15 mmol

Hence, calculate the amount of heat released as follows.

Heat released in the reaction(q) = $m \times s \times \Delta T$

as, m = mass of solution

and, Density = $\frac{mass}{volume}$

or, mass = Density × Volume

= 1.08 g/ml \times (50 + 50) ml

= 108 g

where, s = specific heat of solution = 4.18 j/g.k

and, change in temperature $\Delta T$ = $(26 - 23)^{o}C$

= $3
^{o}C$

Hence, the heat released will be as follows.

q = $m \times s \times \Delta T$

q = $108 \times 4.18 \times 3^{o}C$

= 1354.32 joule

or, = 1.354 kJ (as 1 kJ = 1000 J)

Also, $\Delta H_{rxn}$ = $\frac{-q}{n}$

= $\frac{-1.354}{15 \times 10^{-3}}$

= -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

6 0

2 years ago