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saul85 [17]
2 years 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:
zysi [14]2 years ago
7 0
SAMPLE A - <span>pure substance.
</span>SAMPLE B - <span>homogeneous mixture.
</span>SAMPLE C - <span>heterogeneous mixture.
</span>Pure substance - <span>constant composition and properties.</span>
Homogeneous mixture - same uniform appearance and composition.
Heterogeneous mixture - <span>not </span>uniform<span> in composition, two phases (liquid and dust).
</span>
Marta_Voda [28]2 years ago
4 0

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

Explanation:

Homogeneous mixture : In these mixture, all the component are uniformly distributed throughout the mixture.For example: sugar-water solution.

Heterogeneous mixture : In these mixture, all the component are not uniformly distributed throughout the mixture.For example: sand-water solution.

Pure substance : These are the substance made up only one type of atom or molecule.They have fixed value of boiling and melting points. For example: diamond, pure sugar etc.

Sample A is a clear liquid which evaporates at 70^oC with no change in appearance. Hence, pure substance.

Sample B is a clear blue liquid which boils at 90^oC and leaves blue crystals behind with change in appearance. Hence, homogeneous mixture

Sample C is a opaque, whitish liquid which boils at 100^oC and leaves white crystals behind with dust particles settled in the bottom. Hence, heterogeneous mixture.

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Complete combustion of a compound containing hydrogen and carbon produced 2.641 g of carbon dioxide and 1.442 grams of water as
klio [65]

  The empirical   formula  of hydrocarbon is C₃H₈

The  molecular formula of hydrocarbon   is C₆H₁₆


<u><em> Empirical  formula  calculation</em></u>

Hydrocarbon  contain  carbon and hydrogen

  Step 1:  find the  mass  carbon (C) in carbon dioxide (CO₂)  and hydrogen (H )  in water

mass of  of element = molar mass  of element/ molar mass molecule x total mass of    molecule

From periodic table the molar mass  of C =12,    for CO₂ = 12+( 16 x2) =44 g/mol,     for H = 1.00 g/mol,    for H₂O = (2 x1)+16 = 18 g/mol

mass of C = 12/44 x 2.641 =0.7203 g

since there are 2 atom  of H in H₂O the molar mass of H = 1 x2 = 2 g/mol

mass of H  is therefore =  2/18 x 1.442 =0.1602 g


Step 2:  find the moles of C and H

moles = mass÷ molar mass

moles of C = 0.7203 g÷ 12 g/mol = 0.060  moles

moles of H  =  0.1602÷ 1 g/mol = 0.1602 moles


Step 3: find the mole ratio  of C and H by dividing  each  mole by smallest mole ( 0.06)

for C = 0.06/0.06 =1

  For H = 0.1602/0.06 =2.67

multiply   by 3  to remove the decimal

For C = 1 x3 =3

For H = 2.67 x3 =8

therefore the empirical formula = C₃H₈


<u><em>The molecular formula calculation</em></u>

[C₃H₈]n  = 88.1 g/mol

[12 x 3)+( 1 x8)]n =88.1 g/mol

44 n = 88.1

divide both side by 44

n=2

therefore [C₃H₈]₂   = C₆H₁₆



7 0
1 year ago
Read 2 more answers
The image below shows a laser that has been shot into a solid block of transparent plastic from the far left end. Once inside th
OlgaM077 [116]

Answer:

A. Optical fibers transmit light signals in high-speed communications.  

Explanation:

You didn't include the image, but it probably showed light bouncing off the sides as in the diagram below.

It demonstrates how optical fibers transmit light signals in high-speed communications.

B is wrong. Satellites communicate by radio waves.

C is wrong. Solar cells convert light energy to  

D is wrong. Power plants transmit electrical energy to homes through copper wires.

7 0
2 years ago
You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain
IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of O_2:-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{49.8\ g}{32\ g/mol}

Moles_{O_2}= 1.55625\ mol

Since pressure and volume are constant, we can use the Avogadro's law  as:-

\frac {V_1}{n_1}=\frac {V_2}{n_2}

Given ,  

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

\frac {V_1}{n_1}=\frac {V_2}{n_2}

\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of CO_2 = 44.0 g/mol

Mass of CO_2 = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>

5 0
1 year ago
A solution is prepared by adding 100 mL of 1.0 M HC2H3O2 (aq) to 100 mL of 1.0 M NaC2H3O2 (aq). The solution is stirred and its
DIA [1.3K]

Answer:

(C) H3O+(aq) + C2H3O2−(aq) -> HC2H3O2(aq) + H2O(l)

Explanation:

A buffer is a solution of a weak acid and its salt. It mitigates against changes in acidity or alkalinity of a system. A buffer maintains the pH at a constant value by switching the equilibrium concentration of the conjugate acid or conjugate base respectively.

Addition if an acid shifts the equilibrium position towards the conjugate acid side while addition of a base shifts the equilibrium position towards the conjugate base side.

5 0
2 years ago
Given that at 25.0 ∘C Ka for HCN is 4.9×10−10 and Kb for NH3 is 1.8×10−5, calculate Kb for CN− and Ka for NH4+. Enter the Kb val
neonofarm [45]

Explanation:

Using the expression :

K_a\times K_b=K_w

Where, K_w is the dissociation constant of water.

At 25\ ^0C, K_w=10^{-14}

Thus, for HCN , K_a=4.9\times 10^{-10}

<u>K_b for CN⁻ can be calculated as:</u>

K_a\times K_b=K_w

4.9\times 10^{-10}\times K_b=10^{-14}

K_b=2.0\times 10^{-5}

Thus, for NH₃ , K_b=1.8\times 10^{-5}

<u>K_a for NH_4^+ can be calculated as:</u>

K_a\times K_b=K_w

K_a\times 1.8\times 10^{-5}=10^{-14}

K_a=5.6\times 10^{-10}

5 0
2 years ago
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