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2 years ago

You are trying to determine the

Chemistry

1 answer:

guapka [62]2 years ago

3 0

Answer:

Answer:

Mass = 27.291 g

Explanation:

Given data:

Mass of sodium chloride = ?

Mass of water = 534 g

Molality = 0.875 m

Solution:

Molality:

It is the number of moles of solute into kilogram of solvent.

Formula:

Molality = number of moles of solute / kilogram solvent

Mathematical expression:

m = n/kg

Now we will convert the g into kg.

Mass of water = 534 g× 1kg/1000 g = 0.534 kg

Now we will put the values in formula.

0.875 m = n / 0.534 kg

n = 0.467 mol

Now we will calculate the mass of sodium chloride:

Mass = number of moles × molar mass

Mass = 0.467 mol × 58.44 g/mol

Mass = 27.291 g

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if you desire a slower falling parachute to protect the body from damage,

silk is the best

if you desire a faster falling parachute to escape enemy bullets,

nylon is the best

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If gatorade g series, endurance formula, contains 14 grams of carbohydrate per 8 ounces, how many ounces should a basketball pla

Mashcka [7]

Based on recommended amount of carbohydrate, a basketball player should consume about 17 - 34 ounces of gatorade g series during the hour-long game.

<h3>How many ounces of endurance formula gatorade g series, endurance formula should a basketball player consume during an hour-long game if it contains 14 grams of carbohydrate per 8 ounces?</h3>

Carbohydrates are food substances metabolized easily by the body to produce energy.

Given that the recommended amount of carbohydrate to consume to maintain performance is 30–60 g/h.

Also 14 grams of carbohydrate found in 8 ounces of the drink.

30 g of carbohydrate will be present in 30 × 8/14 = 17.1 ounces of gatorade g series

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How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

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b. Approximately $7.2\; \rm mL$ .

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The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

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Solution in a.:

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Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

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$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

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