A gas has a volume of 300 mL and a pressure of 2 atm. What volume will the gas occupy when the pressure isincreased to 7 atm (total)?

Answers

Answer 1
Answer:

Answer:

The answer is 85.71 mL

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the new volume

V_2 =  (P_1V_1)/(P_2)  \n

We have

V_2 =  (300 * 2)/(7)  =  (600)/(7)   \n  = 85.714285...

We have the final answer as

85.71 mL

Hope this helps you


Related Questions

Use the weather map to answer the question.Weather map with blue line with triangles pointing down and to the right. Red line with semi-circles pointing up and to the right. Red line is to the right of the blue line.© Dorling Kindersley / Universal Images Group / Image Quest 2016What does the red line indicate?A. A cold front is moving to the north and east.B. A cold front is moving to the south and west.C. A warm front is moving to the north and east.D. A warm front is moving to the south and west.
Water contains 2 polar bonds and the molecule is polarTrueFalse​
Write a net ionic equation for the reaction that occurs when excess hydrochloric acid (aq) and potassium sulfite (aq) are combined. Note: Sulfites follow the same solubility trends as sulfates.
Which compound is classified as an electrolyte? a) C6H12O6 b)Ca(OH)2 c)C12H22O11 d)CH3OH
1. Potassium (K) has an atomic mass of 39.0983 amu and only two naturally-occurring isotopes. The K-41 isotope (40.9618 amu) has a natural abundance of 6.7302%. What is the mass (in amu) of the other isotope

Answer True or False for each of the following statements. (a) The carburization surface was maintained at slightly less than 0.25 wt% carbon for each specimen. (b) Comparing the finished specimens at a depth of 0.20 mm, specimen A features the lowest carbon concentration. (c) Comparing the finished specimens as a whole, specimen D features the lowest overall amount of carbon.

Answers

Answer:

verdadero/a

falso/b

verdadero/c

Explanation:

In a calorimetry experiment, it was determined that a 92.0 gram piece of copper metal released 1860 J of heat to the surrounding water in the calorimeter (qcopper = −1860 J). If the final temperature of the copper metal-water mixture was 25.00°C, what was the initial temperature of the copper metal? The specific heat of copper is 0.377 J/(g°C). Group of answer choices Tinitial = 28.6°C Tinitial = −28.6°C Tinitial = 78.6°C Tinitial = 92.6°C Tinitial = 53.6°C

Answers

Answer:

Tinitial = 78.6°C

Explanation:

In a calorimetry experiment, the flow heat is measured for a system that has a state change. In this case, there isn't happening a physical change, so the heat is called sensitive heat, and it's calculated by:

q = mxCpxΔT

Where q is the heat, m is the mass, Cp is the specific heat and ΔT is the difference of final and initial temperature (Tfinal - Tinitial).

Copper is losing heat, so q is negative, then:

-1860 = 92x0.377x(25 - Ti)

34.684(25 - Ti) = -1860

25 - Ti = -53.63

-Ti = -78.63

Ti = 78.6ºC

Predict the sign of the entropy change,Delta S, for each of the following reactions:The signs are either going to be pos or negativea) Pb^2+(aq) + 2Cl-(aq) ---> PbCl2(s)b) CaCO3(s) ---> CaO(s) + CO2 (g)c) 2NH3(g) ---> N2(g) + 3H2(g)d) P4(g) + 5O2(g) ---> P4O10(s)e) C4H8(g) + 6O2(g) ---> 4CO2(g) + 4H2O(g)f) I2(s) ---> I2(g)

Answers

Answer: a) Pb^(2+)(aq)+2Cl^-(aq)\rightarrow PbCl_2(s):  negative

b)  CaCO_3(s)\rightarrow CaO(s)+CO_2(g) : positive

c) 2NH_3(g)\rightarrow N_2(g)+3H_2(g): positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_(10)(s) : negative

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g):  positive.

f) I_2(s)\rightarrow I_2(g) : positive.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

\Delta S is positive when randomness increases and \Delta S is negative when randomness decreases.

a) Pb^(2+)(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)

As ions are moving to solid form , randomness decreases and thus sign of \Delta S is negative.

b)CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

c)2NH_3(g)\rightarrow N_2(g)+3H_2(g)

As 2 moles of reactants are converted to 4 moles of products , randomness increases and thus sign of\Delta S is positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_(10)(s)

As gas is changing to solid, randomness decreases and thus sign of \Delta S is negative.

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)

As 7 moles of reactants are converted to 8 moles of products , randomness increases and thus sign of\Delta S is positive.

f)I_2(s)\rightarrow I_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

10. Predict the mass of nitrogen dioxide produced if 2.30 L of ammonia are allowed to reactwith excess oxygen gas at STP?​

Answers

Answer:

Mass of nitrogen dioxide produced = 4.6 g

Explanation:

Given data:

Volume of ammonia = 2.30 L

Mass of nitrogen dioxide produced = ?

Solution:

Chemical equation:

4NH₃ + 7O₂     →      4NO₂ + 6H₂O

Number of moles of ammonia at STP:

PV = nRT

n = PV/RT

n = 1 atm × 2.30 L / 0.0821 atm.L/K.mol × 273 K

n = 2.30 atm .L / 22.414 atm.L/mol

n = 0.1 mol

Now we will compare the moles of ammonia with nitrogen dioxide from balance chemical equation.

                NH₃            :             NO₂

                 4                :               4

                 0.1             :              0.1

Mass of NO₂:

Mass = number of moles  × molar mass

Mass = 0.1 mol  × 46 g/mol

Mass = 4.6 g

.......state Hess law​

Answers

Hess's Law of Constant Heat Summation (or just Hess's Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.

If sodium arsenite is Na3AsO3, the formula for calcium arsenite would be

Answers

Answer:

Ca₃(AsO₃)₂

Explanation:

Sodium arsenite, with the chemical formula Na₃AsO₃, is formed  by the cation Na⁺ and the anion AsO₃³⁻. For the molecule to be neutral, 3 cations Na⁺ and 1 anion AsO₃³⁻ are required.

Calcium arsenite would be formed by the cation Ca²⁺ and the anion AsO₃³⁻. For the molecule to be neutral, we require 3 cations Ca²⁺ and 2 anions AsO₃³⁻. The resulting chemical formula is Ca₃(AsO₃)₂.