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stira [4]
3 years ago
9

An air mattress is filled with 16.5 moles of air. The air inside the mattress has a temperature of 295 K and a gauge pressure of

3.5 kilopascals. What is the volume of the air mattress?
The volume of the air mattress is__liters.
Chemistry
2 answers:
postnew [5]3 years ago
7 0
PV=nRT
3.5×X=16.5×0.082×295
X= 114 L
The volume of the air mattress is 114 liters.
Ksivusya [100]3 years ago
7 0

Explanation:

According to ideal gas equation, product of pressure and volume equals n times R times T.

Mathematically,         PV = nRT

where       P = pressure

                 V = volume

                 n = number of moles

                 R = gas constant

                 T = temperature

Therefore, it is given that no. of moles is 16.5 mol, pressure is 3.5 kilopascal equals 0.035 atm (as 1 Kpa = 0.01 atm), temperature is 295 K and R = 0.082 LatmK^{-1}mol^{-1}.

Hence, calculate the volume as follows.

                                   PV = nRT

                          0.035 atm \times V = 16.5 mol \times 0.082 L atmK^{-1}mol^{-1} \times 295 K

                                 V = \frac{399.135 L atm}{0.035 atm}

                                     = 11403.85 L

Hence, we can conclude that the volume of the air mattress is 11403.85 liters.

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Determine the moles of H+ reacting with the metal based on the following experimental data. 10.00 mL of 1.00 M HCl solution is a
musickatia [10]

Answer:

Approximately 1.876 \times 10^{-3}\; \rm mol.

Explanation:

Convert both volumes to standard units (that is: liters.)

  • 10.00 \; \rm mL = 10.00 \times 10^{-3}\; \rm L = 1.000 \times 10^{-2}\; \rm L.
  • 27.08 \; \rm mL = 27.08 \times 10^{-3}\; \rm L = 2.708 \times 10^{-2}\; \rm L.

Number of moles of \rm HCl initially present (in the 10.00\; \rm mL solution at 1.00\; \rm M.)

n(\mathrm{HCl}, \, \text{initial}) = \displaystyle c \cdot V = 1.00\times 1.000 \times 10^{-2}= 1.000\times 10^{-2}\; \rm mol.

Number of moles of \rm NaOH from the titration:

n(\mathrm{NaOH}) = c \cdot V = 0.30 \times 2.708 \times 10^{-2} = 8.124 \times 10^{-3}\; \rm mol.

\rm NaOH neutralizes \rm HCl at a 1:1 ratio:

\rm HCl + NaOH \to NaCl + H_2O.

Hence, n(\mathrm{HCl},\, \text{leftover}) = n(\mathrm{NaOH}) = 8.124 \times 10^{-3}\; \rm mol.

\begin{aligned}&n(\mathrm{HCl},\, \text{consumed}) \\ =& n(\mathrm{HCl},\, \text{initial}) - n(\mathrm{HCl},\, \text{leftover}) \\ =& 1.000\times 10^{-2} - 8.124\times 10^{-3} \\ =& 1.876 \times 10^{-3}\; \rm mol\end{aligned}.

3 0
3 years ago
PLEASE HELP: You have been given 0.507 moles of cesium fluoride (CsF), determine the mass in grams of cesium fluoride that you h
Degger [83]

Answer:

C because 77.0 CsF

Explanation:

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Which of the following will increase the rate of disolving?
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Answer:

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The heat of vaporization of a liquid is 84.0 J/g. How many joules of heat would it take to completely vaporize 172 g of this liq
Aliun [14]

Answer:

14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.

Explanation:

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.

During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.

So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.

In this case, being:

  • Q=?
  • m= 172 g
  • L= 84 \frac{J}{g}

and replacing in the expression Q = m*L you get:

Q=172 g*84 \frac{J}{g}

Q=14,448 J

<u><em>14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.</em></u>

5 0
3 years ago
How many grams of neutral salt will be obtained in the reaction of calcium oxide with 200 cm 3 of phosphoric acid solution whose
katen-ka-za [31]

Answer:

9.3 g of Ca3(PO4)2

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

Next, we shall determine the number of mole of H3PO4 present in 200 cm³ of 0.3 mol/dm³ phosphoric acid (H3PO4) solution. This can be obtained as follow:

Molarity of H3PO4 = 0.3 mol/dm³

Volume = 200 cm³ = 200 cm³/1000 = 0.2 dm³

Mole of H3PO4 =?

Molarity = mole /Volume

0.3 = mole of H3PO4 /0.2

Cross multiply

Mole of H3PO4 = 0.3 × 0.2

Mole of H3PO4 = 0.06 mole

Next, we shall determine the number of mole of the salt, Ca3(PO4)2, obtained from the reaction. This can be obtained as shown below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

From the balanced equation above,

2 moles of H3PO4 reacted to produced 1 mole of Ca3(PO4)2.

Therefore, 0.06 moles of H3PO4 will react to produce = (0.06 × 1)/2 = 0.03 mole of Ca3(PO4)2.

Thus, 0.03 mole of Ca3(PO4)2 is produced from the reaction.

Finally, we shall determine the mass of Ca3(PO4)2 produced as follow:

Mole of Ca3(PO4)2 = 0.03 mole

Molar mass of Ca3(PO4)2 = (40×3) + 2[31 + (16×4)]

= 120 + 2[31 + 64]

= 120 + 2[95]

= 120 + 190

= 310 g/mol

Mass of Ca3(PO4)2 =?

Mole = mass /Molar mass

0.03 = mass of Ca3(PO4)2 / 310

Cross multiply

Mass of Ca3(PO4)2 = 0.03 × 310

Mass of Ca3(PO4)2 = 9.3 g

Thus, 9.3 g of Ca3(PO4)2 was obtained from the reaction.

6 0
2 years ago
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