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Blababa [14]
3 years ago
10

Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45

g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon)
Chemistry
1 answer:
musickatia [10]3 years ago
5 0

Answer:

For 1 antacid tablet (in ballon1) we get .0173 moles of CO2

for 2 tablets (in balloon 2) we get: 2*0,0173=  0.0346 moles of CO2

For 3 tablets (in balloon 3) we get 3* 0.0173 = 0.0519 moles of CO2

Explanation:

The complete question:

Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon).

Balloon 1 had 1 antacid tab

Baloon 2 had 2

Balloon 3 had 3

Step 1: Data given

1.45 g of NaHCO3 is present in an antacid tablet

Molar mass of NaHCO3 = 84.00 g/mol

Step 2: The balanced equation

NaHCO3 + H2O → NaOH + H2O + CO2

Step 3: Calculate moles of NaHCO3

Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3

1.45g / 84.0 g/mol = .0173 moles

Step 4: Calculate moles CO2

For 1 mol NaHCO3 we need 1 mol H2O to produce 1 mol NaOH 1 mol H2O and 1 mol CO2

For 0.0173 moles NaHCO3 we'll get 0.0173 moles CO2

so for 1 antacid tablet we get .0173 moles of CO2

for 2 tablets we get: 2*0,0173 =  0.0346 moles of CO2

For 3 tablets we get 3* 0.0173 = 0.0519 moles of CO2

 

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MrRa [10]
<h3>Answer:</h3>

78.34 g

<h3>Explanation:</h3>

From the question we are given;

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<h3>Step 1: Writing the balanced equation for the reaction </h3>

The Balanced equation for the reaction is;

   N₂(g) + 3H₂(g) → 2NH₃(g)

<h3>Step 2: Calculating the number of moles of NH₃</h3>

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ × 2

                                  = 2.3 moles × 2

                                  = 4.6 moles

<h3>Step 3: Calculating the mass of ammonia produced </h3>

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3 0
3 years ago
A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther
sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )

1. Calculate the thermometer readings after  0.5 min and 1 min

(a) After 0.5 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}

(b) After 1 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

  t = t - 1, because the cooling starts 1 min late

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}

3. Plot the temperature readings as a function of time.

The graphs are shown below.

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Answer:

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