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Alik [6]
4 years ago
15

Upon completing a dilution from a stock solution, you will always have more volume of solution than you started with and it will

have a lower concentration than the stock solution.true or false
Chemistry
1 answer:
dlinn [17]4 years ago
3 0
True: due to
<span>- A stock solution is a solution that has been prepared with a greater concentration of solute than is needed for a procedure, so it must be diluted to the proper concentration(s) before use.
- It is very important that you master the concepts and techniques of preparing dilutions .
- A dilution is made by taking a measured portion of a stock solution, and adding to it a solvent (water or a buffer).
- This creates a new solution with a lower concentration of solute than was contained in the stock solution.
- Since the solute concentration of the stock solution is known as well as the volumes of stock and solvent, it is possible to calculate the concentration of solute in the dilute solution.
- The dilute solution is always less concentrated than the stock or starting solution. 
</span>So the answer is True
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Which of the following is a chemical property of iron? It
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Answer:

is capable of combining with oxygen to form iron oxide

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Think about your morning from when you wake up until you arrive at school. Describe the forms of energy you encounter or use dur
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Answer:

When sleeping the energy exhibited is Potential energy. When I wake up and I’m preparing for the day I use kinetic energy which is the energy exhibited during motion.

Eating before leaving the house means I used chemical energy for replenishment of nutrients to aid the performance of more activities for the day. The food eaten and the physical activities also give rise to thermal energy which is why we sweat.

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3 years ago
How much heat is required to raise the temperature of 225 grams of ice from -26.8 °C to steam at 133 °C ?
lara [203]
<h3>Answer:</h3>

150000 J

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Thermodynamics</u>

Specific Heat Formula: q = mcΔT

  • <em>q</em> is heat (in J)
  • <em>m</em> is mass (in g)
  • <em>c</em> is specific heat (in J/g °C)
  • ΔT is change in temperature (in °C or K)

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right
<h3>Explanation:</h3>

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] <em>m</em> = 225 g

[Given] <em>c</em> = 4.184 J/g °C

[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C

[Solve] <em>q</em>

<u>Step 2: Solve for </u><em><u>q</u></em>

  1. Substitute in variables [Specific Heat Formula]:                                          q = (225 g)(4.184 J/g °C)(159.8 °C)
  2. Multiply:                                                                                                           q = (941.4 J/°C)(159.8 °C)
  3. Multiply:                                                                                                           q = 150436 J

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

150436 J ≈ 150000 J

Topic: AP Chemistry

Unit: Thermodynamics

Book: Pearson AP Chemistry

5 0
3 years ago
When someone’s immune system overreacts to something harmless it’s called:
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5 0
4 years ago
Read 2 more answers
A 41.1 g sample of solid CO2 (dry ice) is added to a container at a temperature of 100 K with a volume of 3.4 L.A. If the contai
marta [7]

Answer:

Approximately 6.81 × 10⁵ Pa.

Assumption: carbon dioxide behaves like an ideal gas.

Explanation:

Look up the relative atomic mass of carbon and oxygen on a modern periodic table:

  • C: 12.011;
  • O: 15.999.

Calculate the molar mass of carbon dioxide \rm CO_2:

M\!\left(\mathrm{CO_2}\right) = 12.011 + 2\times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

Find the number of moles of molecules in that 41.1\;\rm g sample of \rm CO_2:

n = \dfrac{m}{M} = \dfrac{41.1}{44.009} \approx 0.933900\; \rm mol.

If carbon dioxide behaves like an ideal gas, it should satisfy the ideal gas equation when it is inside a container:

P \cdot V = n \cdot R \cdot T,

where

  • P is the pressure inside the container.
  • V is the volume of the container.
  • n is the number of moles of particles (molecules, or atoms in case of noble gases) in the gas.
  • R is the ideal gas constant.
  • T is the absolute temperature of the gas.

Rearrange the equation to find an expression for P, the pressure inside the container.

\displaystyle P = \frac{n \cdot R \cdot T}{V}.

Look up the ideal gas constant in the appropriate units.

R = 8.314 \times 10^3\; \rm L \cdot Pa \cdot K^{-1} \cdot mol^{-1}.

Evaluate the expression for P:

\begin{aligned} P &=\rm \frac{0.933900\; mol \times 8.314 \times 10^3 \; L \cdot Pa \cdot K^{-1} \cdot mol^{-1} \times 298\; K}{3.4\; L} \cr &\approx \rm 6.81\times 10^5\; Pa \end{aligned}.

Apply dimensional analysis to verify the unit of pressure.

4 0
4 years ago
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