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STALIN [3.7K]
3 years ago
10

How much NaCl would you dissolve and dilute to the 100.00 mL mark in a volumetric flask with DI H2O to prepare a 0.825 M sodium

chloride solution?
Chemistry
1 answer:
adoni [48]3 years ago
4 0

Answer:

4.82 g

Explanation:

To solve this problem we'll use the <em>definition of molarity</em>:

  • Molarity = moles / liters

We are given the volume and concentration (keep in mind that 100mL=0.100L):

  • 0.825 M = moles / 0.100 L

Now we <u>calculate the number of NaCl moles required</u>:

  • moles = 0.0825 mol

Then we convert 0.0825 NaCl moles into grams, using its molar mass:

  • 0.0825 mol * 58.44 g/mol = 4.82 g
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Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0
olya-2409 [2.1K]

Answer:

pH=-1.37

Explanation:

We are given that 25 mL of 0.10 M CH_3COOH is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of CH_3COOH=25mL=0.025 L

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL=\frac{35}{1000}=0.035 L

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of CH_3COOH=0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+=0.10\times 0.025=0.0025 moles

Number of moles of OH^-=0.10\times 0.01=0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+=\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L

pH=-log [H+]=-log [4.28\times 10^{-2}]=-log4.28+2 log 10=-0.631+2

pH=-1.37

6 0
3 years ago
The National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. The units the NWS
pashok25 [27]

Answer:

d. 103.3

Explanation:

In the given question, the National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. And the units of atmospheric pressure used for reporting the atmospheric pressure data are inches of mercury. For a barometric pressure of 30.51 inches of mercury, we can calculate the pressure in kPa as follow:

In principle, 3.386 kPa is equivalent to the atmospheric pressure of 1 inch of mercury. Thus, 30.51 inches of mercury is equivalent to 30.51 in *(3.386 kPa/1 in) = 103.307 kPa.

Therefore, a barometric pressure of 30.51 inches of mercury corresponds to _____103.3_____ kPa.

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Try putting this in biology category not chemistry so more ppl can help :)
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The units of ppm means parts per million. Also, It is equivalent to milligrams per liter. It is one way of expressing concentration of a substance. It u<span>sually used to describe the concentration of something in water or soil. We calculate the mass of CaCO3 as follows:

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A(n) _____ is a region where fresh water and salt water mix.
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