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3 years ago

Water circulates through Earth's water cycle by changing state state its acidity its acidity its turbidity

Chemistry

1 answer:

Lana71 [14]3 years ago

3 0

Complete question is;

Water circulates through Earth's water cycle by changing its _____.

A. state

B. acidity

C. turbidity

D. composition

Answer:

A: State

Explanation:

We can talk about this by observations about the states of water in the Earth's water cycle and even nature.

For example, water changing from liquid to steam(gas) in evaporation; under certain temperatures/conditions snow undergoes sublimation directly from solid to gas and also undergoes melting from solid to liquid; water vapour in the atmosphere undergoes condensation and cooling to form rain.

Thus, we can say that water circulates through the Earth's water by changing it's state.

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Give an example of chemical equilibrium you have come across into your day-to-day life. Explain how it meets the definition of e

Roman55 [17]

There are many examples of chemical equilibrium all around you. One example is a bottle of fizzy cooldrink. In the bottle there is carbon dioxide (CO2) dissolved in the liquid. There is also CO2 gas in the space between the liquid and the cap

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2 years ago

+ c) FeCl3 + NH4OH Fe(OH)3 NHACI

bixtya [17]

The question is incomplete, the complete question is:

Write the net ionic equation for the below chemical reaction:

(c): $FeCl_3+3NH_4OH\rightarrow Fe(OH)_3+3NH_4CI$

Answer: The net ionic equation is $Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)$

Explanation:

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

(c):

The balanced molecular equation is:

$FeCl_3(aq)+3NH_4OH(aq)\rightarrow Fe(OH)_3(s)+3NH_4Cl(aq)$

The complete ionic equation follows:

$Fe^{3+}(aq)+3Cl^-(aq)+3NH_4^+(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)+3NH_4^+(aq)+3Cl^-(aq)$

As ammonium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)$

5 0

3 years ago

During nuclear fission and fusion, what is matter that seems to disappear actually converted into?

hjlf

Answer:

During nuclear fission and fusion matter that seems to disappear but is actually converted into energy. The amount of energy (E) produced in such a reaction can be calculated using Einstein's formula for the equivalence of mass and energy: E = mc^2.

Explanation:

4 0

3 years ago

Read 2 more answers

a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: Moles solution = moles solute + moles solvent. First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$