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Ugo
1 month ago
13

Simon has collected three samples from the coral reef where he observes marine life. He must determine whether each one is a pur

e substance or a mixture.
SAMPLE A : Clear liquid. Evaporates at 70*C. Appearance does not change.
SAMPLE B : Clear, blue liquid. Boils at 90*C leaving blue crystals behind. Appearance does not change.
SAMPLE C : Opaque, whitish liquid. Boils at 100*C leaving white crystals behind. Dust appears to settle at bottom.

Answer choices for each one are - homogeneous mixture, heterogeneous mixture, or pure substance
Chemistry
2 answers:
lorasvet [2.7K]1 month ago
7 0
SAMPLE A - <span>pure substance.
</span>SAMPLE B - <span>homogeneous mixture.
</span>SAMPLE C - <span>heterogeneous mixture.
</span>Pure substance - <span>has a consistent composition and distinct properties.</span>
Homogeneous mixture - exhibits a uniform look and composition throughout.
Heterogeneous mixture - <span>is </span>not uniform<span> in its composition, containing two distinct phases (liquid and particulate matter).
</span>
Tems11 [2.7K]1 month ago
4 0

Answer: Sample A is classified as a pure substance, Sample B is identified as a homogeneous mixture, and Sample C is a heterogeneous mixture.

Explanation:

Homogeneous mixture: In these mixtures, all components are evenly distributed. For example: a sugar-water solution.

Heterogeneous mixture: In these mixtures, components are not uniformly distributed. For instance: a sand-water solution.

Pure substance: These are made from a single type of atom or molecule and are characterized by fixed boiling and melting points. Examples include diamond and pure sugar.

Sample A is a clear liquid evaporating at 70^oC with a consistent appearance. Thus, it is a pure substance.

Sample B is a clear blue liquid boiling at 90^oC that leaves behind blue crystals, resulting in a change in appearance. Hence, it is a homogeneous mixture

Sample C is an opaque, whitish liquid boiling at 100^oC, leaving white crystals, with dust settling at the bottom. Thus, it is a heterogeneous mixture.

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Compound A reacts with Compound B to form only one product, Compound C, and it's known the usual percent yield of C in this reac
VMariaS [2998]

Response:

1. 16.54 grams.

2. 6.64 grams.

Clarification:

Greetings,

In this situation, the chemical reaction taking place is:

A+B\rightarrow C

Thus, based on the provided data, the theoretical yield of C can be determined as:

m_C^{theoretical}=\frac{12.9g}{0.78}=16.54g

Additionally, taking into account the principle of conservation of mass, the mass prior to the reaction is equal to the mass following the reaction, hence, the mass of B that was utilized amounts to:

m_B=m_C-m_A=16.54g-10.0g=6.64g

Best regards.

6 0
1 month ago
A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T
KiRa [2933]

Answer:

\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}

Explanation:

1. Molar concentration

Designate chloroform as C and acetone as A.

The molar concentration for C is derived from Moles of C per Litres of solution.

(a) Moles of C

We are assuming there are 0.187 moles of C.

This resolves that step.

(b) Litres of solution

Next, identify 0.813 moles of A.

(i) Mass of each component

\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}

(ii) Volume of each component

\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}

(iii) Volume of solution

Assuming mixing doesn't alter the total volume.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}

2. Molal concentration of C

Molal concentration is calculated as moles of solute per kilograms of solvent.

Total moles of C = 0.187 mol.

Mass of A = 47.22 g = 0.047 22 kg.

\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}

4 0
2 months ago
No amino acid molecule by itself can speed up or catalyze reactions between other molecules; however, when amino acids are joine
Anarel [2989]

Answer:

This indicates that the enzyme is a type of protein.

Explanation:

It is important to remember that proteins are composed of vast numbers of amino acids. Because these amino acids are tiny units, they cannot function as a catalyst on their own.

However, when they form a polymer, the protein enzyme will possess varying shapes, sizes, and both physical and chemical attributes differing from a single monomer.

Additionally, for proteins to function actively, a specific number of amino acids must combine to create a distinct shape suited to interact with another molecule, thus accelerating the chemical reaction and functioning as an enzyme.

8 0
2 months ago
Read 2 more answers
Cathodic protection of a metal pipe against corrosion usually entails __________.
VMariaS [2998]

Answer:

C. connecting an active metal to designate the pipe as the cathode in an electrochemical cell.

Explanation:

Cathodic protection involves a method to manage the accelerated corrosion of a metal surface by designating it as the cathode within an electrochemical cell. This is accomplished by attaching the protected metal to a more sacrificial metal, which acts as the anode.

This method helps to preserve the metal by introducing a highly reactive metal that serves as the anode, supplying free electrons. By adding these free electrons, the active metal gives up its ions, protecting the less reactive steel from corrosion.

3 0
2 months ago
How much CO2 (L) is produced when 2.10 kg of sodium bicarbonate reacts with excess hydrochloric acid at 25.0 °C and 1.23 atm? A)
KiRa [2933]

The equation representing the reaction between sodium bicarbonate and hydrochloric acid is as follows:

NaHCO_3_(_s_) + HCl_(_a_q_) \implies NaCl_(_a_q_) + CO_2_(_g_) + H_2O_(_l_)

The substances NaHCO_3 and HCl combine in a 1:1 ratio. Therefore, we calculate the quantity of sodium bicarbonate and its molar mass to determine the moles formed.

NaHCO_3_M_r = 22.99 + 1.008 + 12.011+ 3 \times 16.0= 84.01 g/mol.

2.1kg\ NaHCO_3 \times \frac{1000g}{kg} \times \frac{mol}{84.01g/mol} = 24.997\ mol.

We also recognize that the stoichiometric proportions are 1:1:1:1:1, which leads to the conclusion that the moles of CO_2 equal 24.977 moles.

Next, we apply the ideal gas equation PV=nRT, where P denotes pressure, V refers to volume, R is the gas constant, and T represents the temperature in kelvins. We rearrange to solve for V

PV= nRT \implies V= \frac{nRT}{P}= \frac{ 24.997\ mol \times 8.2507m^3\ atm \times 298.15K }{mol \times K \times 1.23 atm} = 49967\ m^3

The final answer should be expressed in liters, 1L = 1000\ m^3, hence

49967\ m^3 \times\frac{L}{1000\ m^3} =49.97L\ CO_2\ produced

6 0
2 months ago
Read 2 more answers
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