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navik [9.2K]
3 years ago
10

A 500 mL piston has a gas mixture inside at a pressure of 15.0 atm. What volume would the piston have if the pressure increased

to 30.0 atm?
Chemistry
1 answer:
spayn [35]3 years ago
6 0

Answer:

The volume of the piston would be 250 mL.

Explanation:

Volume of the piston, V_1=500\ mL

Pressure, P_1=15\ atm

Let V_2 is the volume of piston if the pressure increased to 30.0 atm.

It is based on Boyle's law. It is given by:

\dfrac{V_1}{V_2}=\dfrac{P_2}{P_1}\\\\V_2=\dfrac{V_1P_1}{P_2}\\\\V_2=\dfrac{500\times 15}{30}\\\\V_2=250\ mL

So, the volume of the piston would be 250 mL.

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Convert the following to Fahrenheit 10 Celsius 30 Celsius 40 Celsius 37 Celsius 0 Celsius
ipn [44]
10 = 50
30 = 86
40 = 104
37 = 98.6
0 = 32
5 0
3 years ago
Read 2 more answers
If you combine 290.0 mL 290.0 mL of water at 25.00 ∘ C 25.00 ∘C and 140.0 mL 140.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t
Alexxandr [17]

Answer:

The final temperature is 47.79 °C

Explanation:

Step 1: Data given

Sample 1 has a volume of 290.0 mL

Temperature of sample 1 = 25.00 °C

Sample 2 has a volume of 140.00 mL

Temperature of sample 2 = 95.00 °C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qlost = -Qgained

Q = m*c* ΔT

Q(sample1) = -Q(sample2)

m(sample1) * c(sample1) * ΔT(sample1) = -m(sample2)*c(sample2) *ΔT(sample2)

⇒with m(sample1) = the mass of sample 1 = 290.0 mL * 1g/mL = 290 grams

⇒with c(sample 1) = the specific heat of water = c(sample 2)

⇒with ΔT(sample 1) = the change of temperature = T2 - T1 = T2 - 25.00 °C

⇒with m(sample2) = the mass of sample 2 = 140.0 mL * 1g/mL = 140 grams

⇒with c(sample2) = the specific heat of water = c(sample1)

⇒with ΔT(sample2) = T2 -T1 = T2 - 95.00°C

m(sample1) *  ΔT(sample1) = -m(sample2)*ΔT(sample2)

290 *(T2-25.0) = -140 *(T2 - 95.0)

290 T2 - 7250 = -140 T2 + 13300

430 T2 = 20550

T2 = 47.79 °C

The final temperature is 47.79 °C

4 0
3 years ago
On the first day of your new job as a chemist, you are given a bottle of magnesium sulfate and asked to make 30 mL of 0.3 M MgSO
Lady bird [3.3K]

Answer:

The answer to your question is 15500 mg of MgSO₄ 7H₂O

Explanation:

Data

Volume = 30 ml

Concentration = 0.3 M

Formula MgSO4∗7H2O

Process

1.- Calculate the number of moles needed

 Molarity = moles / volume

Solve for moles

 moles = Molarity x volume

Substitution

 moles = 0.3 x 0.03 l

 moles = 0.009 moles of MgSO₄

2.- Calculate the molecular mass of MgSO4∗7H2O

molecular mass = 24 + 32 + 64 + 14 + 112

                           = 246 g

3.- Get the proportion MgSO4 :7H2O, this proportion is 1 : 7

4.- Calculate the amount of MgSO4∗7H2O

    0.009 mol of MgSO4 \frac{7 MgSO4 7H2O}{1 mol MgSO4} \frac{246 g MgSO4 7H2O}{1 mol MgSO4 7H2O}

Simplification

   15.5 g = 15500 mg

5 0
3 years ago
Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:
Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

                                   \frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} ),

where  λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from n_1 to n_2 and R_h = 109677 is the Rydberg Constant. Here n_1 and  n_2 represents the transitions.

(a) n_1 =2 to n_2 = infinity

            \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2}  ) = 109677/4     [since 1/infinity = 0] Therefore, \lambda = 4 / 109677 = 0.00003647 m

(b) n_1=4 to  n_2 = 20

           \frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2}  ) = 6580.62

Therefore,  \lambda = 1 / 6580.62 = 0.000152 m

(c) n_1=3 to  n_2 = 10

          \frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2}  ) = 11089.56

Therefore,  \lambda = 1 / 11089.56 = 0.00009 m

(d)  n_1=2 to  n_2 = 1

          \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2}  ) = - 82257.75

Therefore,  \lambda = 1 /82257.75  = - 0.0000121 m  

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

#SPJ4

8 0
1 year ago
What tests can you do to identify gases, based on their chemical properties?
GuDViN [60]
Tests for gases
Hydrogen, oxygen, carbon dioxide, ammonia and chlorine can be identified using different tests.
Hydrogen. A lighted wooden splint makes a popping sound in a test tube of hydrogen.
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