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tatyana61 [14]
3 years ago
14

Liquid water in the earth's air is an example of a solution. what type of solution would this be?

Chemistry
1 answer:
murzikaleks [220]3 years ago
6 0
Solutions are homogeneous mixtures made up of two non-reacting compounds. The component which is present in large quantity is called as solvent and the component in small quantity is called as solute There are six types of solutions.
                                       
1. Gas (solvent) - Gas (solute) Example= Air

2. Liquid (solvent) - Gas (solute) Example= Fizzy Drinks

3. Liquid (solvent) - Liquid (solute) Example= 96% Ethanol in Water

4. Liquid (solvent) - Solid (solute) Example= Sugar in Tea

5. Solid (solvent) - Solid (solute) Example= Brass, 18 caret Gold

6. Solid (solvent) - Gas (solute) Example= Marshmellow

Result:
          In equation earth air is solvent (as it is in large quantity) and water vapors (gas) which are in small quantity is solute.
So, it is Gas - Gas Solution.
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The lattice energy of a salt is related to the energy required to separate the ions. For which of the following pairs of ions is
kati45 [8]

Answer:

(A) Mg²⁺ and O²⁻

Explanation:

6 0
2 years ago
Calculate the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL.
Finger [1]

Considering the definition of dilution, the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL is 0.93 mL.

<h3>What is diluion</h3>

First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

  • Ci: initial concentration
  • Vi: initial volume
  • Cf: final concentration
  • Vf: final volume

<h3>Molarity of the solution in this case</h3>

In this case, you know:

  • Ci= 4 M
  • Vi= 7 mL
  • Cf= ?
  • Vf= 30 mL

Replacing in the definition of dilution:

4 M× 7 mL= Cf× 30 mL

Solving:

(4 M× 7 mL)÷ 30 mL= Cf

<u><em>Cf= 0.93 mL</em></u>

In summary, the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL is 0.93 mL.

Learn more about dilution:

brainly.com/question/20113402

brainly.com/question/22762236

8 0
2 years ago
Answer questions based on the lab activity.
Alika [10]

Answer:

Conduction, Convection and Conduction

Explanation:

8 0
3 years ago
When C2H6(g) reacts with O2(g) according to the following reaction, 1.43×103 kJ of energy are evolved for each mole of C2H6(g) t
dezoksy [38]

Answer:

-2.86x10³ kJ

Explanation:

The enthalpy of a reaction (ΔH) is defined as the heat produced or consumed by a reaction. In the reaction:

2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(g)

The ΔH is the heat envolved in the reaction per 2 moles of C₂H₆. 1.43x10³ kJ are involved when 1 mole reacts. Thus, when 2 moles react, involved heat is:

1.43x10³ kJ ₓ 2 = <em>2.86x10³ kJ</em>. As the reaction is a combustion reaction (Produce CO₂ and H₂O), the heat involved in the reaction is <em>PRODUCED, </em>that means ΔH is negative, <em>-2.86x10³ kJ</em>

8 0
3 years ago
Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE
nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let "\alpha is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

                  HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}

Initial                  C                         0            0

Equilibrium        c(1- \alpha)              c\alpha c\alpha

Dissociation constant = Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}

=\frac{c\alpha^{2}}{(1-\alpha)}

In this case weak acids \alpha is very small so, (1-\alpha ) is taken as 1.

Ka=C\alpha^{2}

\alpha=\sqrt\frac{ka}{c}

From the given the concentration = 0.050 M

Substitute the given value.

\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}

[H_{3}O^{+}]=c\alpha

[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}

pH= -log[H_{3}O^{+}]

=-log[4.9\times10^{-6}]

=6-log 4.9= 5.31

Therefore, The pH of the solution is 5.31.

7 0
3 years ago
Read 2 more answers
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