CHEMISTRY COLLEGE

Question 2 Here are four sketches of pure substances. Each sketch is drawn as if a sample of the substance were under a microscope so powerful that individual atoms could be seen. Decide whether each sketch shows a sample of an element, a compound, or a mixture.

Answers

Answer 1

Answer:

The image with the four sketches that accompany this question was obtained in internet and is attached.

Answer:

Substance X: compound

Substance Y: mixture

Substance W: element

Substance Z: compound

Explanation:

The first picture shows substance X as a matter composed by particles of one type. Every particle consists of two white balls bonded to one red ball, sketching a compound formed by two "white atoms" and one "red atom". So, this sketch shows a sample of a compound.

The second picture shows substance Y formed by two different kind of particles. One type of particles are two white balls bonded, sketching a molecule of two equal atoms, i.e. an element in the form of a diatomic molecules of a single element (likely a gas). The other type of particles are two white balls bonded to one red ball (just as in the first picture). This sketch, then, shows the mixture of one element and one compound. Thus, this is a mixture.

The third picture shows substance W formed by one type of particles: groups of two identical white balls. So, every pair of white ball sketchs a diatomic molecule, which means that this is an element (think, for example, in H₂ or O₂).

The fourth picture shows substance Z formed by groups of particles of the same type: two red balls bonded to two white balls. So, every group of these particles represents identical compounds, making this a compound and not a mixture.

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Aqueous solutions of sodium hypoch lorite (NaOCI), best known as bleach, are prepared by the reaction of sodium hydroxide with chlorine: 2 NaOH (aq)Cl2(g)->NaOCI (aq)+ H20 (I)+ NaCl (aq) How many grams of NaOH are needed to react with 25.0 g of chlorine?

Answers

Answer:

28.2 g of NaOH

Explanation:

We need to calculate the grams of NaOH needed to react with 25.0 g of Cl₂ in the following reaction:

2 NaOH(aq) + Cl₂(g) → NaOCI(aq + H₂0(I) + NaCl(aq)

We are going to solve this by making use of the molar ratio between Cl₂ and NaOH given by the reaction equation where we see that every mol of Cl₂ will react with 2 moles of NaOH.

So first we need to convert the 25.0 g of Cl₂ to moles:

Molar Mass of Cl₂ = 2 x 35.45 = 70.90 g/mol
Moles of Cl₂ = 25.0 g / 70.90 g/mol = 0.3526 moles

Then we need to calculate the moles of NaOH needed to react with these moles of Cl₂ knowing that every mol of Cl₂ will react with 2 moles of NaOH:

moles of NaOH = 2 x moles of Cl₂ = 2 x 0.3526 moles = 0.7052 moles

Next we must convert these moles to grams:

Molar Mass of NaOH = 22.990 + 15.999 + 1.008 = 40.00 g/mol
Mass of NaOH = 0.7052 moles x 40.00 g/mol = 28.2 g

28.2 g are needed to react with 25.0 g of Cl₂ in the production of NaOCl

Consider two aqueous solutions of NaCl. Solution 1 is 4.00 M and solution 2 is 0.10 M. In what ratio (solution 1 to solution 2) must these solutions be mixed in order to produce a 0.86 M solution of NaCl

Answers

Answer:

The ratio of solution 1 to solution 2 is 24.20 to 100.00.

Explanation:

We will mix V₁ (L) of solution 1 with V₂ (L) of solution 2 to get the final solution.

So the mole concentration in the final solution is calculated as below, note that C₁ is the concentration of solution 1, and C₂ is the concentration of solution 2

$[M] = (V_(1) C_(1) +V_(2)C_(2))/(V_(1)+V_(2)) = \frac{4 V_(1) + 0.1 V_(2)}_{V_(1)+V_(2)}}=0.86$

Then we can calculate for the ratio

$(V_(1))/(V_(2))=(0.86-0.10)/(4.00-0.86) =(0.76)/(3.14)$ or $(24.20)/(100.00)$

Briefly explain why sublimation occurs.

Answers

Answer:

Sublimation is basically cause by the heat absorption and this process is an endothermic since it require extra energy.

It basically provide sufficient energy to molecules to control the various type of attractive forces from the neighbors and then convert it into the vapor phase.

Sublimation occur when the particle of gases become cold because some substances has high vapor pressure. Sublimation is the endothermic change and it occur below the triple point in terms of pressure and temperature.

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first ignoring ionic strength and activities. a. silver iodate
b. barium sulfate
c. Repeat the above calculations using ionic strength and activities.

Answers

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_(sp) =\text{[Ag$^(+)$][IO$_(3)$$^(-)$]} = s* s = s^(2) = 3.0* 10^(-8)\ns = \sqrt{3.0* 10^(-8)} \text{ mol/L} = 1.7 * 10^(-4) \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_(sp) =\text{[Ba$^(2+)$][SO$_(4)$$^(2-)$]} = (0.02 + s) * s \approx 0.02s = 1.1* 10^(-10)\ns = (1.1* 10^(-10))/(0.02) \text{ mol/L} = 5.5 * 10^(-9) \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = (1)/(2) \sum_(i) {c_(i)z_(i)^(2)}\n\n\mu = (1)/(2) (\text{[Ba$^(2+)$]}\cdot (2+)^(2) + \text{[NO$_(3)$$^(-)$]}*(-1)^(2)) = (1)/(2) (\text{0.02}* 4 + \text{0.04}*1)= (1)/(2) (0.08 + 0.04)\n\n= (1)/(2) *0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(1)^(2)√(0.06) = -0.51* 0.24 = -0.12\n\gamma = 10^(-0.12) = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_(sp) =\text{[Ag$^(+)$]$\gamma_(Ag^(+))$[IO$_(3)$$^(-)$]$\gamma_{IO_(3)^(-)}$} = s*0.75* s * 0.75 =0.56s^(2)= 3.0 * 10^(-8)\ns^(2) = (3.0 * 10^(-8))/(0.56) = 5.3 * 10^(-8)\n\ns =2.3 * 10^(-4)\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(2)^(2)√(0.06) = -0.51*16* 0.24 = -0.50\n\gamma = 10^(-0.50) = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_(sp) =\text{[Ba$^(2+)$]$\gamma_( Ba^(2+))$[SO$_(4)$$^(2-)$]$\gamma_{ SO_(4)^(2-)}$} = (0.02 + s) * 0.32* s* 0.32 \approx 0.02*0.10s\n2.0* 10^(-3)s = 1.1 * 10^(-10)\ns = (1.1* 10^(-10))/(2.0 * 10^(-3)) \text{ mol/L} = 5.5 * 10^(-8) \text{ mol/L}$

Fireworks explode above your
head. Is this exothermic or
endothermic and why?

Answers

I think it is exothermic

Identify each element below, and give the symbols of the other elements in its group. a. [Ar] 4s23d104p4
b. [Xe] 6s24f145d2
c. [Ar] 4s23d5.

Answers

Answer:

Explanation:

the electron configuration is defined as the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It is

used to describe the orbitals of an atom in its ground state

The valence electrons, electrons in the outermost shell, can be used to know the chemical property

Chemical Name of the Element: Selenium

Chemical Symbol: Se

Group it belong in periodic table:6A

Other Element in the same group:tellurium(Te),,sulfur(S)

atomic number = 34

Selenium is a chemical element that has symbol Se It is a nonmetal which is usually classified as metalloid with properties that are intermediate between the elements above and below in the periodic table.

b)Chemical Name of the Element:Hafnium

Chemical Symbol: Hf

Group it belong in periodic table:4B

Other Element in the same group: Titanium( Ti )Rutherfordium

atomic number: 72

Hafnium is a solid at room temperature.

c)Chemical Name of the Element: Manganese

Chemical Symbol:Mg

Group it belong in periodic table:Mn

Other Element in the same group:Bohrium(Bh) ,Technetium(Tc)

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