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defon
3 years ago
12

Which methods will help you properly identify pH with indicator paper? Check all that apply. Set a piece of pH paper in front of

each solution before testing to keep organized. Dip the pH paper into each test tube and lay it on the table to let it change color. Use a transfer pipet to remove a few drops from the solution to drop onto the paper. Wait a minute or so before reading color. Use the same piece of indicator paper for each solution to conserve paper.
Chemistry
2 answers:
AnnyKZ [126]3 years ago
7 0

Answer:

A) Set a piece of pH paper in front of each solution before testing to keep organized.

C) Use a transfer pipet to remove a few drops from the solution to drop onto the paper.

D) Wait a minute or so before reading color.

monitta3 years ago
4 0

Set a piece of pH paper in front of each solution before testing to keep organized.

Use a transfer pipet to remove a few drops from the solution to drop onto the paper.

Wait a minute or so before reading color.

Is the correct answer...

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What is a neutron and what charge does it have
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It’s a polyatomic and it has a negative charge. It’s located in the nucleus of the atom, along with the protons.
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Differentiate between short period and long period in modern periodic table​
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The difference between short period and long period is based upon the number of elements in each period. Shortest period is the first period which contains elements, while the longest period is the 6th period which contains 32 elements.
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For alpha decay to cause a loss in mass, such as that in the uranium 238 conversion to thorium 234, what must the atomic weight
Aleonysh [2.5K]

Explanation:

Reaction equation showing alpha decay in Uranium-238 is as follows.

    ^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He

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Symbol of an alpha particle is ^{4}_{2}\alpha.

As atomic mass or weight is the sum of total number of protons and neutrons present in an atom.

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6 0
2 years ago
If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction
Tasya [4]

Answer:

<u></u>

  • <u>0.456M</u>

Explanation:

<u>1. Balanced molecular equation</u>

     2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O

<u>2. Mole ratio</u>

     \dfrac{2molHNO_3}{1molBa(OH)_2}

<u>3. Moles of HNO₃</u>

  • Number of moles = Molarity × Volume in liters
  • n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

<u>4. Moles Ba(OH)₂</u>

  • n = 0.700M × 0.0310 liter = 0.0217 mol

<u>5. Limiting reactant</u>

Actual ratio:

   \dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

<u>6. Final molarity of Ba(OH)₂</u>

  • Molarity = number of moles / volume in liters
  • Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M
5 0
3 years ago
The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of
timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = 1.83g/cm^3=1.83g/ml

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml

Now we have to calculate the molarity of solution.

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0
3 years ago
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