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lidiya [134]
3 years ago
8

A common anesthetic in dentistry known as nitronox consists of a mixture of dinitrogen oxide, n2o, and oxygen gas, o2, which is

administered through an inhaler over the nose. If the anesthetic mixture has a total pressure of 740 mmhg, and the partial pressure of the oxygen is 370 mmhg, what is the partial pressure, in torr, of the dinitrogen oxide?
Chemistry
1 answer:
Nesterboy [21]3 years ago
3 0

Anesthetic in dentistry  consists of a mixture of dinitrogen oxide (N₂O) and oxygen gas (O₂), which is administered through an inhaler over the nose. Total pressure of the mixture (P_{total}) is sum of partial pressure of N₂O (p_{N_{2}O }) and partial pressure of O₂, (p_{O_{2} }).

P_{total}=p_{N_{2}O } + p_{O_{2} }

P_{total}= 740 mmHg, p_{O_{2} }=370 mmHg

So, p_{N_{2}O }= (740-370) mmHg= 370 mmHg=370 torr

Hence, partial pressure of N₂O is 370 torr.

As, 1mmHg= 1 torr.


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Is chlorine a substance or a mixture
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Answer:

its a substance

Explanation:

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2. Aluminum & Chlorine: AlCl3
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5 0
2 years ago
A silver cube with an edge length of 2.38 cm and a gold cube with an edge length of 2.79 cm are both heated to 81.9 ∘C and place
never [62]

Answer:

The final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

Explanation:

<u>Given data;</u>

edge length of silver, a = 2.38 cm = 0.0238 m

edge length of gold, a = 2.79 cm = 0.0279 m

final temperature of silver, t = 81.9 ° C

final temperature of Gold, t = 81.9 ° C

initial temperature of water, t = 19.6 ° C

volume of water, v =  109.5 mL = 0.0001095 m³

<u>Known data:</u>

density gold 19300 kg/m³

density silver 10490 kg/m³

density water 1000 kg/m³

specific heat gold is 129 J/kgC

specific heat silver is 240 J/kgC

specific heat water is 4200 J/kgC

<u>Calculated data</u>

Apply Pythagoras theorem to determine the side of each cube;

Silver cube;

let L be the side of the silver cube

Taking the cross section of the cube (form a right angled triangle), the edge  length forms the <em>hypotenuse side</em>.

L² + L² = 0.0238²

2L² = 0.0238²

L² = 0.0238² / 2

L² = 0.00028322

L = √0.00028322

L = 0.0168

Volume of cube = L³

Volume of the silver cube = (0.0168)³ = 4.742 x 10⁻⁶ m³

Gold cube;

let L be the side of the gold cube

L² + L² = 0.0279²

2L² = 0.0279²

L² = 0.0279² / 2

L² = 0.0003892

L = √0.0003892

L = 0.0197

Volume of cube = L³

Volume of the silver cube = (0.0197)³ = 7.645 x 10⁻⁶ m³

Mass of silver cube;

density = mass / volume

mass = density x volume

mass of silver cube = 10490 (kg/m³) x 4.742 x 10⁻⁶ (m³) = 0.0497 kg

Mass of Gold cube

mass of gold cube = 19300 (kg/m³) x 7.645 x 10⁻⁶ (m³) = 0.148 kg

Mass of water

mass of water = 1000 (kg/m³) x 0.0001095 (m³) = 0.1095 kg

Let the heat gained by cold water be  Q₁

Let the heat lost  by silver cube = Q₂

Let the heat lost  by gold cube = Q₃

Let the final temperature of water = T

Q₁  = 0.1095 kg x 4200 J/kgC x (T–19.6)

Q₂ = 0.0497 kg x 240 J/kgC x (81.9–T)

Q₃ = 0.148 kg x 129 J/kgC x (81.9–T)

At thermal equilibrium;

Q₁ = Q₂ + Q₃

0.1095 x 4200  (T–19.6)  = 0.0497 x 240 x (81.9–T)  + 0.148 x 129 x (81.9–T)

459.9 (T–19.6) = 11.928 (81.9–T) + 19.092(81.9–T)

459.9T - 9014.04 = 976.9032 - 11.928T + 1563.6348 - 19.092T

459.9T + 11.928T + 19.092T = 9014.04  + 976.9032 + 1563.6348

490.92T = 11554.578

T = 11554.578 / 490.92

T = 23.54 ⁰C

Therefore, the final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

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3 years ago
A battery can provide a current of 4.60 A at 3.40 V for 2.50 hr. How much energy (in kJ) is produced? 1st attempt kJ Energy
Lunna [17]

Answer:

The energy produced equals 140.760 kJ

Explanation:

The relation between power, current and voltage is

Power=Current\times Voltage

Applying the given values in the relation above we get

Power=4.60\times 3.40=15.64W

Now Since Power=\frac{Energy}{Time}\\\\Energy=Power\times Time

Again applying the calculated values we get

Energy=15.64\times 2.50\times 3600=140760Joules=140.76kJ

4 0
3 years ago
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
3 years ago
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