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3 years ago

A gas occupied 23.55 L at 118.2 Celsius. At what temp would the gas occupied 43.8 L

Chemistry

1 answer:

pantera1 [17]3 years ago

4 0

Answer:

The gas occupies 43.8 L at a temperature of 219.8 °C

Explanation:

<u>Step 1:</u> Data given

Volume of the gas = 23.55L

Temperature = 118.2 °C

New volume = 43.8 L

<u>Step 2</u>: Calculate changed temperature

(P1*V1)/T1 = (P2*V2)/T2

Let's assume the pressure stays constant, then we have:

V1/T1 = V2/T2

⇒ with V1 = the initial volume = 23.55L

⇒ with T1 = the initial temperature = 118.2 °C

⇒ with V2 = the changed volume = 43.8 L

⇒ with T2 = the changed temperature = TO BE DETERMINED

23.55L/118.2°C = 43.8 L / T2

T2 = 219.8 °C

The gas occupies 43.8 L at a temperature of 219.8 °C

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Compute the molar enthalpy of combustion of glucose (C6 H12O6 ): C6 H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (g) Given that com

lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)

Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c(calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 *6.30 + 755 * 6.30 = 4764.5 J ( negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

5 0

3 years ago

How many moles do you need to prepare 1.675 L of a 8.50 M NaOH solution? Round your answer to 2 decimal places.

vampirchik [111]

Answer:

0.20 mol's

Explanation:

1.675 L = 1.675 dm^3

moles = V/(conc):

moles = 1.675/(8.5)

moles = 0.1970... --> 0.20

7 0

3 years ago

Read 2 more answers

You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho

timama [110]

Explanation:

As per the given data, at a higher temperature, at $24^{o}C$ , the solution will occupy a larger volume than at $20^{o}C$ .

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At $20^{o}C$ , density of water=0.9982071 g/ml

Therefore, $\frac{concentration}{density}$ will be calculated as follows.

= $\frac{C_{1}}{d_{1}}$

= $\frac{1.000 mol/L}{0.9982071 g/ml}$

= 1.0017961 mol/g

At $24^{o}C$ , density of water = 0.9972995 g/ml

Since, $\frac{concentration}{density}$ = $\frac{C_{2}}{d_{2}}$

= $\frac{C_{2}}{0.9972995}$

Also, $\frac{C_{1}}{d_{1}}$ = $\frac{C_{2}}{d_{2}}$

so, 1.0017961 mol/g = $\frac{C_{2}}{0.9972995}$

$C_{2} = 1.0017961 \times 0.9972995$

= 0.9990907 mol/L

Therefore, in 500 ml, concentration of $KNO_{3}$ present is calculated as follows.

$C_{2}$ = $\frac{concentration}{volume}$

0.9990907 mol/L = $\frac{concentration}{0.5 L}$

concentration = 0.49954537 mol

Hence, mass (m'') = $0.49954537 mol \times 101.1032 g/mol$ = 50.5056 g (as molar mass of $KNO_{3}$ = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

m = $m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}$

where, $d_{air}$ = density of air = 0.0012 g/ml

$d_{weight}$ = density of callibration weights = 8.0g/ml

d = density of weighed object

Hence, the true mass will be calculated as follows.

True mass(m) = $50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}$

true mass(m) = 50.5268 g

= 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of $KNO_{3}$ needs to be measured.

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3 years ago

A taco from a fast food restaurant contains 7.0 g of fat, 25. g of carbohydrate, and 11 g of protein. How many kJ of energy does

Gennadij [26K]

Answer: 878 kJ of energy is there in cheeseburger

Explanation:

Given : calorific value of fat = 38kJ/g

calorific value of carbohydrate = 17 kJ/g

calorific value of protein = 17kJ/g

1 g of fat contains energy = 38 kJ

7.0 g of fat contains energy = $\frac{38}{1}\times 7.0=266 kJ$

1 g of carbohydrate contains energy = 17 kJ

25 g of fat contains energy = $\frac{17}{1}\times 25=425kJ$

1 g of protein contains energy = 17 kJ

11 g of protein contains energy = $\frac{17}{1}\times 11=187kJ$

Total energy = (266+425+187) kJ = 878 kJ

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3 years ago

Is graph paper a homogeneous or heterogeneous mixture

labwork [276]

Answer:

There is no chemical symbol for paper since it is not an element but rather a mixture of several different compounds. Mixtures in two or more phases are heterogeneous mixtures.

Explanation:

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3 years ago