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2 years ago

approximately how many grams of sodium chloride (NaCl) are required to prepare 500. mL of 3.00 M solution

1 answer:

8 0

First convert mL to L:
(500. mL)/(1000L) = .500L

Then convert molarity to moles:
(3.00M)(.500L) = 1.5mol NaCl

Then convert moles of NaCl to grams of NaCl by multiplying it by its molar mass:
(1.5mol NaCl)(58.44g/mol NaCl) = 87.66g NaCl are needed

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It's this one $100, 5,000

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2 years ago

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4 Identify Give an example of a physical change and an example of a chemical change.

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Answer:

Answer below

Explanation:

Physical changes dont change the chemical makeup of an object. An example would be to tear a piece of paper. Tearing the paper doesnt change how the paper was made chemically

Chemical changes on the other hand do change the chemical makeup of an object. An example of a chemical change would be to bake a cake. The reason being that the chemical makeup of the cake batter is changed so that the actual cake is formed.

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2 years ago

If the boron nitride molecule, bn, were to form, what would its structure look like? add bonds and lone pairs as necessary.

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Answer: The structure for the boron nitride molecule will be linear.

Explanation: When boron and nitrogen combine together to form a compound, the boron element will loose 3 electrons which will be easily accepted by the nitrogen element.

Boron : Atomic number is 5

Electronic configuration: $1s^22s^22p^1$

To attain stable configuration, boron will loose 3 electrons easily.

Nitrogen : Atomic number is 7

Electronic configuration: $is^22s^22p^3$

To attain stable configuration. nitrogen will gain 3 electrons easily.

Hence, to form a molecule with these two elements, there will be 3 bonds present in between them and 1 pair of lone pair will be left on the nitrogen atom.

Structure of boron nitride ( which is linear) is attached in the image below.

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3 years ago

Calculate the rms speed of co molecules at 320 k .

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<span>rms = (3RT/M)^1/2 Therefore, the rms of CO is: rms(CO) = [(3)(8.3145 J/mol*K)(320 K)/(0.028 kg/mol)]^1/2 rms = [2.65x10^5 m^2/s^2]^1/2 rms = 533.9181 m/s</span>

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3 years ago

An 8.42 g piece of metal with a specific heat of 1.020 J g-1 k-1 wasn't heated to an unknown temperature. The metal was then pla

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Answer:

$T = 61.344\,^{\textdegree}C$

Explanation:

The heat received by water is equal to the heat rejected by the piece of metal. That is to say:

$-Q_{piece} = Q_{water}$

$(8.42\,g) \cdot \left(1.020\,\frac{J}{g\cdot ^{\textdegree}C} \right) \cdot (T - 20.40\,^{\textdegree}C)= (44.2\,g)\cdot \left(4.187\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (20.40\,^{\textdegree}C-18.50\,^{\textdegree}C)$

$\left(8.588\,\frac{J}{^{\textdegree}C} \right)\cdot (T-20.40\,^{\textdegree}C) = 351.624\,J$

The initial temperature of the piece of metal is:

$T = 61.344\,^{\textdegree}C$

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3 years ago

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