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Margarita [4]
3 years ago
14

How is this effect of dissolvedsubstances important in nature?​

Chemistry
1 answer:
dimaraw [331]3 years ago
8 0

Explanation:

Water is called the "universal solvent" because it is capable of dissolving more substances than any other liquid. This is important to every living thing on earth. It means that wherever water goes, either through the air, the ground, or through our bodies, it takes along valuable chemicals, minerals, and nutrients.

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A calorie is another term for carbohydrate.TF
Hunter-Best [27]
<span>This is false. A carbohydrate is a carbon-based molecule that can be utilized by living organisms in order to produce energy. A calorie is a unit of energy often used to measure the amount of energy within food. Another example of energy unit is the Joule, more commonly used within physics.</span>
8 0
3 years ago
Explain why chemistry affects all aspects of life and most natural events
Alecsey [184]

Chemistry affects all aspects of life and most natural events because all living and nonliving things are made of matter.

6 0
3 years ago
a 25.0-ml volume of a sodium hydroxide solution requires 19.6 ml of a 0.189 m hydrochloric acid for neutralization. a 10.0- ml v
Rashid [163]

<u>Concentration of NaOH = 0.148 molar, M</u>

<u>Concentration of H3PO4 = 0.172 molar, M</u>

<u></u>

Concentration x Volume  will give the number of moles of solute in that volume.  C*V = moles

Concentration  has a unit of (moles/liter).  When multiplied by the liters of solution used, the result is the number of moles.

Original HCl solution:  (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl

The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl.  The reaction is:

  NaOH + HCl > NaCl and H2O

This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each.  We just determined that the  moles of HCl used was 0.00370 moles.  Therefore, the 25.0 ml solution of NaOH had the same number of moles:  0.00370 moles NaOH.

The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters).  The concentration of NaOH is therefore:  

    <u>(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M</u>

====

The phosphoric acid problem is handled the same way, but with an added twist.  Phosphoric acid is H3PO4.  We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4.  But now the acid has three hydrogens that will react.  The balanced equation for this reaction is:

  H3PO4 + 3NaOH = Na3PO4 + 3H2O

Now we need <u><em>three times</em></u> the moles of NaOH to neutralize 1 mole of H3PO4.

The moles of NaOH that were used is:

  (0.148M)*(0.0349 liters) = 0.00517 moles of NaOH

Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.

The 0.00172 moles of H3PO4 was contained in 10.0 ml.  The concentration is therefore:

     (0.00172 moles H3PO4)/(0.010 liters H3PO4)

<u>Concentration of H3PO4 = 0.172 molar, M</u>

 

5 0
1 year ago
How many grams of KCl is needed to make .75 L of a 1 M solution of KCl?
Zielflug [23.3K]

Answer:

The answer to your question is 25.9 g of KCl

Explanation:

Data

Grams of KCl = ?

Volume = 0.75 l

Molarity = 1 M

Formula

Molarity = \frac{number of moles}{volume}

Solve for number of moles

Number of moles = Molarity x volume

Substitution

Number of moles = 1 x 0.75

Simplification

Number of moles = 0.75 moles

Molecular mass KCl = 39 + 35.5 = 34.5

Use proportions to find the grams of KCl

                          34.5 g of KCl ----------------  1 mol

                             x                  ----------------  0.75 moles

                            x = (0.75 x 34.5) / 1

                            x = 25.9 g of KCl

3 0
3 years ago
? Answer the question below. Type your response in the space provided. What volume of a 2.5 M stock solution of acetic acid (HC2
Novay_Z [31]
Data: 
M_{concentrated} = 2.5\:mol
V_{concentrated} = ?
M_{dilute} = 0.50\:mol
V_{dilute} = 100\:mL\to0.100\:L
<span>
Formula: Dilution Calculations

</span>M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}
<span>
Solving:

</span>
M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}
2.5 * V_{concentrated} = 0.50 * 0.100
2.5V_{concentrated} = 0.05
V_{concentrated} =  \frac{0.05}{2.5}
\boxed{\boxed{V_{concentrated} = 0.02\:L\:or\:20\:mL}} \end{array}}\qquad\quad\checkmark<span>







</span>
3 0
3 years ago
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