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Igoryamba
3 years ago
8

Waters states of matter include steam liquid water and ice. What about water is the same in the states? What can you conclude ab

out what changes and what does not change during a change of state?
Chemistry
1 answer:
julia-pushkina [17]3 years ago
5 0
This lesson is the first in a three-part series that addresses a concept that is central to the understanding of the water cycle—that water is able to take many forms but is still water. This series of lessons is designed to prepare students to understand that most substances may exist as solids, liquids, or gases depending on the temperature, pressure, and nature of that substance. This knowledge is critical to understanding that water in our world is constantly cycling as a solid, liquid, or gas.

In these lessons, students will observe, measure, and describe water as it changes state. It is important to note that students at this level "...should become familiar with the freezing of water and melting of ice (with no change in weight), the disappearance of wetness into the air, and the appearance of water on cold surfaces. Evaporation and condensation will mean nothing different from disappearance and appearance, perhaps for several years, until students begin to understand that the evaporated water is still present in the form of invisibly small molecules." (Benchmarks for Science Literacy<span>, </span>pp. 66-67.)

In this lesson, students explore how water can change from a solid to a liquid and then back again.

<span>In </span>Water 2: Disappearing Water, students will focus on the concept that water can go back and forth from one form to another and the amount of water will remain the same.

Water 3: Melting and Freezing<span> allows students to investigate what happens to the amount of different substances as they change from a solid to a liquid or a liquid to a solid.</span>
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What is the net ionic equation for the reaction if any that occurs when aqueous solutions of Na2CO3 and HCL are mixed?
strojnjashka [21]

Answer:

CO₃²⁻(aq)  +  2H⁺(aq) → CO₂ (g)  + H₂O (l)

Explanation:

The balanced reaction between Na2CO3 and HCl is given as;

Na₂CO₃ (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO₂ (g) + H₂O (l)

The next step is o express the species as ions.

The complete ionic equation for the above  reaction would be;

2Na⁺(aq)  + CO₃²⁻(aq)  +  2H⁺(aq)  + 2Cl⁻(aq)   → Na⁺(aq)  + Cl⁻(aq)  + CO₂ (g)  + H₂O (l)

The next step is to cancel out the spectator ion ions; that is the ions that appear in both the reactant and product side unchanged.

The spectator ions are;  Na⁺ and Cl⁻

The net ionic equation is given as;

CO₃²⁻(aq)  +  2H⁺(aq) → CO₂ (g)  + H₂O (l)

7 0
3 years ago
How does the law of definite proportions apply to hydrates?
vladimir1956 [14]
The law of definite proportions would state that a hydrate always contain exactly the same proportion of salt and water by mass.
strictly speaking, the law of definite proportion states that a compound always 
contains exactly the same proportion of elements by mass.
But the law is often applied to groupings of elements in compound.
Hydrates are salt that have a certain amount of  water asa part of their structure.
The water is chemically combined with the compound in a definite ratio.
4 0
3 years ago
Sulphur burnt in oxygen forms sulphur dioxide. Calculate the heat given out when 1.6g of sulphur is burnt.
adell [148]

Answer:

Dont know

Explanation:

4 0
2 years ago
An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g
vodomira [7]

Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity= \frac{33.8 g solute}{100 g solution} x \frac{1.15 g solution}{1 ml} x \frac{1 mol solute}{145.6 g solute} x \frac{1000 ml}{1 L}

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x \frac{1 mol solute}{145.6 g} = 0.232 mol

mol H20= 66.2 g H₂O x \frac{1 mol H2O}{18 g}

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute=\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}

Xsolute= 0.059

4 0
2 years ago
A 30.5 g sample of an alloy at 95.0°C is placed into 49.3 g water at 24.3°C in an insulated coffee cup. The heat capacity of the
user100 [1]

Answer:

0.752 J/g*K

Explanation:

The heat lost by the alloy (which is negative) must be equal to the heat gained by the water and the coffee cup:

-Qa = Qw + Qc

-ma*ca*ΔTa = mw*cw*ΔTw + C*ΔTc

Where, m is the mass, c is the specific heat capacity, C is the heat capacity of the coffee cup, ΔT is the change in temperature, a represents the alloy, and w the water.

The coffee cup has initial temperature equal to the water, then:

-30.5*ca*(31.1 - 95.0) = 49.3*4.184*(31.1 - 24.3) + 9.2*(31.1 - 24.3)

1948.95ca = 1465.20

ca = 0.752 J/g*K

4 0
3 years ago
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