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tatyana61 [14]
2 years ago
15

Use Boyle's Law to interpret the following problem and then select the correct answer below.

Chemistry
1 answer:
FromTheMoon [43]2 years ago
4 0

The pressure of the compressed tank that was initially filled with hydrogen gas and has a pressure of 100kPa is pressure in the 250mL tank increases.

<h3>What is Boyle's law?</h3>

Boyle's law states that the observation that the pressure of an ideal gas is inversely proportional to its volume at constant temperature.

According to this question, a 1 L compressed tank is filled with hydrogen gas and has a pressure of 100kPa.

If all of the hydrogen gas is removed from the tank and pumped into a 250mL tank, which is smaller in volume, the pressure in the 250mL tank increases.

Learn more about pressure at: brainly.com/question/12971272

#SPJ1

You might be interested in
A gas sample occupies 200 mL at 760 mm Hg. What volume does the gas occupy at 400 mm Hg?
astra-53 [7]

Answer:

201.9

Explanation:

when you divided 760 with 400 yo get 19.0 the add it with 200 you get that answer

3 0
3 years ago
One liter of air contains about 0.21 L of oxygen. When filled, the human lungs hold about 6.0 L of air. How much oxygen is in th
Pie
Simple 6L of air times the .21L of O2 = 1.26

3 0
3 years ago
A truck tire has a volume of 218 L and is filled with air to 35.0 psi at 295 K. After a drive, the air heats up to 318 K. (b) If
Alenkasestr [34]

The pressure (in psi) is 37.1

We are given the following information:

- The initial temperature of the air, $T_{1}=295 \mathrm{~K}$

- The initial pressure, $P_{1}=35.0 \mathrm{psi}$.

- The initial volume, $V_{1}=208 \mathrm{~L}$

- The final temperature, $T_{2}=319 \mathrm{~K}$

- The increase in the volume is $2 \%$, that is, $\Delta V=\frac{2}{100} V_{1}$ where $\Delta V$is the increase in the volume.

The combined gas law states that a fixed amount of an ideal gas obeys the following equation: $\frac{P V}{T}=$ constant, where:

- P is the Pressure of the gas.

- V is the Volume of the gas.

- n is the number of moles of gas.

$R=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}=0.0821 \mathrm{~L} \cdot {atm} / \mathrm{mol} \cdot \mathrm{K}$ is the Universal Gas constant.

- T is the absolute temperature of the gas.

The final volume of the air is:

$$\begin{aligned}V_{2} &=V_{1}+\Delta V \\&=V_{1}+\frac{2}{100} V_{1} \\&=\frac{102}{100} V_{1}\end{aligned}$$

Equating the initial and final state, we have:

$$\begin{aligned}\frac{P_{2} V_{2}}{T_{2}} &=\frac{P_{1} V_{1}}{T_{1}} \\\Rightarrow P_{2} &=\frac{V_{1}}{V_{2}} \times \frac{T_{2}}{T_{1}} \times P_{1} \\&=\frac{V_{1}}{102 V_{1} / 100} \times \frac{319 \mathrm{~K}}{295 \mathrm{~K}} \times 35.0 \mathrm{psi} \\& \approx 37.1 \mathrm{psi}\end{aligned}$$

The ideal gas law states that a universal constant for an ideal gas is the ratio of the product of pressure and temperature to the product of the number of moles and absolute temperature. The resultant equation is known as the combined gas law if the number of moles in the ideal gas law is set to a constant.

Learn more about combined gas law brainly.com/question/13154969

#SPJ4

8 0
2 years ago
When cobalt chloride is added to pure water, the co2 ions hydrate?
mrs_skeptik [129]
Answer : The question is incomplete and the complete question is attached in the answer;

Answer 1) When HCl was added to the system of cobalt chloride it will eventually produce more Cl ions in the solutions which will shift the equilibrium to the right and make the solution look more blue than before. To accommodate extra Cl ions in the system the equilibrium gets shifted. 

Answer 2) When in the reaction system water is added the Cobalt chloride compound forms a hydrated mixture; this means the equilibrium reaction shifts to the left and produces more hydrated ions along with more chlorine ions in the solution.

Which then turns the resultant solution tinto pink in colour.

3 0
3 years ago
Read 2 more answers
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).
Degger [83]
a) before addition of any KOH : 

when we use the Ka equation & Ka = 4 x 10^-8 : 

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

           = 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

       = 9.2 x 10^-5 M

when PH = -㏒[H+]

   PH = -㏒(9.2 x 10^-5)

        = 4  

b)After addition of 25 mL of KOH: this produces a buffer solution 

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]


first, we have to get moles of HCIO= molarity * volume

                                                           =0.21M * 0.05L

                                                           = 0.0105 moles

then, moles of KOH = molarity * volume 

                                  = 0.21 * 0.025

                                  =0.00525 moles 

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L =  0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

                                        = 0.00525 / 0.075

                                        =0.07 M

and molarity of KCIO = moles KCIO / total volume

                                    = 0.00525 / 0.075

                                    = 0.07 M

and when Ka = 4 x 10^-8 

∴Pka =-㏒Ka

         = -㏒(4 x 10^-8)

         = 7.4 

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4 

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume 

                                                        = 0.21 M * 0.05L

                                                        = 0.0105 moles

then moles KOH = molarity * volume
                            =  0.22 M* 0.035 L 

                            =0.0077 moles 

∴ moles of HCIO remaining = 0.0105 - 0.0077=  8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume 

                                      = 8 x 10^-5 / 0.085

                                      = 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

                                          = 0.0077M / 0.085L

                                          = 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

                                   = 0.0105mol / 0.1 L

                                   = 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

       = 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO] 

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

           = 3.8
∴PH = 14- POH

        =14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

 when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M 

V2 is the excess volume added  of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

           = -㏒0.02 

           = 1.7

∴PH = 14- POH

       = 14- 1.7 

      = 12.3 
8 0
3 years ago
Read 2 more answers
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