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3 years ago

Which material is the least likely to be recognized as a mixture by looking at it under a microscope

Chemistry

2 answers:

PSYCHO15rus [73]3 years ago

5 0

A homogenous mixture is uniform and thus hard to recognize as a mixture. An example is water!

alexgriva [62]3 years ago

5 0

Answer:

C. a homogeneous mixture

Explanation:

I did this on the ed test

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If a 250mL beaker weighs 13.473g and the same beaker plus 2.2 mL of water weighs 15.346g. How much does the 2.2 mL of water, alo

Sindrei [870]

Given :

A 250 ml beaker weighs 13.473 g .

The same beaker plus 2.2 ml of water weighs 15.346 g.

To Find :

How much does the 2.2 ml of water, alone, weigh .

Solution :

Now, mass of water is given by :

$\text{Mass of water alone = Total mass - Mass of beaker alone}\\\\Mass =15.346-13.473\ g\\\\Mass =1.873\ g$

Therefore , mass of 2.2 ml of water alone is 1.873 g .

Hence , this is the required solution .

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3 years ago

What would be the best way to measure the volume of a small ?

WITCHER [35]

You can use a graduated cylinder.

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3 years ago

Ocean basins will expand as new oceanic crust forms and moves away from a mid-oceanic ridge during

KATRIN_1 [288]

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3 years ago

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A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t

soldi70 [24.7K]

Answer:

The pressure in atm calculated using the van der Waals' equation, is 337.2atm

Explanation:

This is the Van der Waals equation for real gases:

(P + a/v² ) ( v-b) = R .T

where P is pressure

v is Volume/mol

R is the gas constant and T, T° in K

a y b are constant for each gas, so those values are data, from the statement.

[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

[P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 24.34 L.atm/mol

0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

0.04181L/mol. P = 14.1 L.atm/mol

P = 14.1 L.atm/mol / 0.04181 mol/L

P = 337.2 atm

4 0

3 years ago

Balance the following redox reaction in acidic solution. Zn(s)+MnO−4(aq)→ Zn+2(aq)+Mn+2(aq)

devlian [24]

Answer:

2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺

Explanation:

To balance a redox reaction in an acidic medium, we simply follow some rules:

Split the reaction into an oxidation and reduction half.
By inspecting, balance the half equations with respect to the charges and atoms.
In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
Multiply both equations with appropriate factors to balance the electrons in the two half equations.
Add up the balanced half equations and cancel out any specie that occur on both sides.
Check to see if the charge and atoms are balanced.

Solution

Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺

The half equations:

Zn → Zn²⁺ Oxidation half

MnO₄⁻ → Mn²⁺ Reduction half

Balancing of atoms(in acidic medium)

Zn → Zn²⁺

MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Balancing of charge

Zn → Zn²⁺ + 2e⁻

MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O

Balancing of electrons

Multiply the oxidation half by 5 and reduction half by 2:

5Zn → 5Zn²⁺ + 10e⁻

2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O

Adding up the two equations gives:

5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O

The net equation gives:

5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O

8 0

3 years ago