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andreev551 [17]
3 years ago
15

Which of the following would likely form a heterogeneous mixture? Sugar and waterSalt and waterSand and waterBaking soda and wat

er
Chemistry
2 answers:
kati45 [8]3 years ago
6 0

Answer:

Sand and water

Explanation:

meriva3 years ago
4 0

Answer:

Water and sand

Explanation:

the prefix hetero- means 'different' so a heterogeneous mixture would be a mixture in which you can clearly see all of the components.

(like a salad)

sugar and baking soda would dissolve in the water (t<em>hat would be an example of a </em><em>homogeneous</em><em> mixture, the prefix homo- means 'same' homogeneous mixtures have a uniform appearance throughout)</em>

However the sand will definitely not dissolve in the water, it will simply sink to the bottom, and stay there. you would clearly be able to see both the water and the sand.

So the answer can be, none other than

Water and sand.

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Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M
kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

3 0
3 years ago
The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.
scZoUnD [109]

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg


Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

5 0
3 years ago
Why is elemental zinc considered a pure substance?
Veseljchak [2.6K]

Answer:  C... Because Zinc is made of only one type of atoms.

Explanation: All elements are made of one type of atom. Pure gold is made out of only gold atoms, silver out of only silver atoms, etc.

7 0
3 years ago
What is the freezing point of a solution of 498mL of water (solute) dissolved in 2.50 L of ethanol (solvent), C2H5OH? The densit
jok3333 [9.3K]

Answer:

Freezing T° of solution is -142.4°C

Explanation:

This excersise is about colligative properties, in this case freezing point depression,

ΔT = Kf . m . i

Where ΔT = Freezing T° of solvent - Freezing T° of solution

Kf = Cryoscopic constant

m = mol/kg (molality)

i = Number of ions dissolved.

Water is not ionic, so i = 1

Let's find out m.

We determine mass of water, by density

498ml . 1 g/mL = 498 g

We convert the mass of water to moles → 498 g . 1mol/18g = 27.6 moles

We determine mass of solvent by density

2500 mL . 0.789 g/mL = 1972.5 g

Notice, we had to convert L to mL to cancel units.

1 cm³ = 1 mL

We convert the mass from g to kg → 1972.5 g . 1kg /1000g = 1.97kg

We determine m = mol/kg → 27.6mol / 1.97kg = 13.9 m

Kf for ethanol is: 1.99 °C/m

Freezing T° for ethanol is: -114.6°C

We replace at formula: - 114.6°C - Freezing T° solution = 1.99 °C/m . 13.9 m . 1

- 114.6°C - Freezing T° solution = 27.8 °C

- Freezing T° solution  = 27.8°C + 114.6°C

Freezing T° Solution = - 142.4 °C

7 0
2 years ago
An iron(iii) sulfate hydrate is 18.4% water. What is the formula of the hydrate? What is the name of the hydrate?
aleksley [76]

Answer:- Formula of the hydrate is Fe_2(SO_4)_3.5H_2O and it's name is Iron(III)sulfate pentahydrate.

Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.

Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.

We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:

81.6gFe_2(SO_4)_3(\frac{1mol}{399.88g})

= 0.204molFe_2(SO_4)_3

18.4gH_2O(\frac{1mol}{18.02g})

= 1.02molH_2O

Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.

Fe_2(SO_4)_3=\frac{0.204}{0.204}  = 1

H_2O=\frac{1.02}{0.204} = 5

There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is Fe_2(SO_4)_3.5H_2O and the name of the hydrate is Iron(III)sulfate pentahydrate.


3 0
2 years ago
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