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Part 1. A chemist reacted 15.0 liters of F2 gas with NaCl in the laboratory to form Cl2 and NaF. Use the ideal gas law equation

xeze [42]

Answer:

113 g NaCl

Explanation:

The Ideal Gas Law equation is:

PV = nRT

In this equation,

> P = pressure (atm)

> V = volume (L)

> n = number of moles

> R = 8.314 (constant)

> T = temperature (K)

The given values all have to due with the conditions fo F₂. You have been given values for all of the variables but moles F₂. Therefore, to find moles F₂, plug each of the values into the Ideal Gas Law equation and simplify.

(1.50 atm)(15.0 L) = n(8.314)(280. K)

2250 = n(2327.92)

0.967 moles F₂ = n

Using the Ideal Gas Law, we determined that the moles of F₂ is 0.967 moles. Now, to find the mass of NaCl that can react with F₂, you need to (1) convert moles F₂ to moles NaCl (via the mole-to-mole ratio using the reaction coefficients) and then (2) convert moles NaCl to grams NaCl (via molar mass from periodic table). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator).

1 F₂ + 2 NaCl ---> Cl₂ + 2NaF

Molar Mass (NaCl): 22.99 g/mol + 35.45 g/mol

Molar Mass (NaCl): 58.44 g/mol

0.967 moles F₂ 2 moles NaCl 58.44 g
---------------------- x ----------------------- x ----------------------- = 113 g NaCl
1 mole F₂ 1 mole NaCl

4 0

2 years ago

HELP PLEASE I GIVE 21 POINTS!!!! +BRAINLIEST!!!!!!!

cricket20 [7]

You could write about a fear you may have had as a child, eg. the dark, or underneath the bed, etc...

7 0

3 years ago

Read 2 more answers

Millions of years ago in many different regions of the world, large-scale burials of organic matter were followed by the formati

kramer

The answer is A: Areas where the geologic process occurred now have major petroleum reserves

5 0

3 years ago

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A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?

4vir4ik [10]

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL, we find the volume of water:

$d = \frac{g}{mL} \\\\ V= \frac{g}{d} = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L$

Now, find moles of $CH_{3} OH$ are needed using the molarity equation:

$M = \frac{ moles }{ V (L)} \\\\\\molesCH_{3}OH = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_{3}OH = 0,375 mol$

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M

5 0

3 years ago

A client who weighs 70 kg is receiving a solution of 0.9% sodium chloride (normal saline) 500 ml with dopamine 800 mg at 5 ml/ho

nordsb [41]

The mcg/kg/minute the client is receiving is 1.904 mcg/kg/min whose weight is 70 kg.

1 milligram (mg) = 1000 micrograms (mcg) or 0.001 grams (g) 1 g = 1000 mg 1 kilogram (kg) = 1000 g 1 kg = 2.2 pound (lb) 1 liter (L) = 1000 milliliters (mL)

The number of mcg/kg/minute = flow rate × concentration ÷ mass of client

Flow rate = 12 ml/hour = 12 ml/hour × 1 hr/60 min = 0.2 ml/min

Concentration = mass of dopamine/volume

where

mass of dopamine = 800 mg and

volume = 500 ml

Concentration =800 mg/500 ml

= 1.6 mg/ml

= 1.6 mg/ml × 1000 mcg/mg

= 1600 mcg/ml

mass of client = 70 kg

Calculating mcg/kg/minute

So, substituting the variables into the equation, we have

mcg/kg/minute = flow rate × concentration ÷ mass of client

mcg/kg/minute = 0.08 ml/min × 1600 mcg/ml ÷ 70 kg

mcg/kg/minute = 320 mcg/min ÷ 70 kg

mcg/kg/minute = 1.904 mcg/kg/min

Thus, the mcg/kg/minute the client is receiving is 1.904 mcg/kg/min

Learn more about mcg/kg/minute here:

brainly.com/question/4253005

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7 0

2 years ago