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anyanavicka [17]
3 years ago
10

Plzzzzzzzzzzzz help me

Chemistry
1 answer:
dedylja [7]3 years ago
7 0

Answer:

its TRUE

Explanation:

These plates lie on top of a partially molten layer of rock called the asthenosphere. Due to the convection of the asthenosphere and lithosphere, the plates move relative to each other at different rates, from two to 15 centimeters (one to six inches) per yea

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Which of the following gas samples would contain the same amount of gas as 200 mL of helium, He(g), at 25° C and 1 atm?
monitta

Answer:

200\; \rm mL of neon \rm Ne\, (g) at 25^{\circ} \rm C and 1\; \rm atm (the second choice) would contain an equal number of gas particles as 200\; \rm mL\! of \rm He \, (g) at 25^{\circ} \rm C\! and 1\; \rm atm\! (assuming that all four gas samples behave like ideal gases.)

Explanation:

By Avogadro's Law, if the temperature and pressure of two ideal gases is the same, the number of gas particles in each gas would be proportional to the volume of that gas.

All four gas samples in this question share the same temperature and pressure. Hence, if all these gases are ideal gases, the number of gas particles in each sample would be proportional to the volume of that sample. Two of these samples would contain the same number of gas particles if and only if the volume of the two samples is equal to one another.

The second choice, 200\; \rm mL of neon \rm Ne\, (g) at 25^{\circ} \rm C and 1\; \rm atm, is the only choice where the volume of the sample is also 200\; \rm mL \!. Hence, that choice would be the only one with as many gas particles as 200\; \rm mL\! of \rm He \, (g) at 25^{\circ} \rm C\! and 1\; \rm atm\!.

7 0
3 years ago
In order to predict the outcome of the reaction, write the molecular, full ionic, and net ionic equations for a mixture of aqueo
Elina [12.6K]

Answer:

See explanation

Explanation:

Full molecular equation;

2NH3(aq) + AgNO3(aq) -------> [Ag(NH3)2]NO3(aq)

Full ionic equation

2NH3(aq) + Ag^+(aq) + NO3^-(aq) --------> [Ag(NH3)2]^+(aq) + NO3^-(aq)

Net ionic equation;

2NH3(aq) + Ag^+(aq) -------->  [Ag(NH3)2]^+(aq)

When Silver nitrate is mixed with a solution of aqueous ammonia, a white and cloudy solution was observed.

6 0
3 years ago
i need help i have a homework assignment to finish and this is the last question help please and thank u
Jet001 [13]
A
i just searched it up and wanted to help
5 0
3 years ago
The sp of pbbr2 is 6. 60×10−6. what is the molar solubility of pbbr2 in pure water?
Inessa05 [86]

Ksp of PbBr₂ is 6.60 × 10⁻⁶. The molar solubility of PbBr₂ in pure water is 0.0118M.

Ksp or Solubility Product Constant is an equilibrium constant for the dissociation in an aqueous solution.

Molar solubility (S) is the concentration of the dissolved substance in a solution that is saturated.

Let the molar solubility be S upon dissociation.

PbBr₂ or Lead Bromide dissociates in pure water as follows:

                          PbBr₂ ----------> Pb⁺² + Br⁻

                                                     S      2S

Ksp = [Pb⁺²] [ Br⁻]

Ksp = (S) (2S)²

Ksp = 4S³

6.60 × 10⁻⁶ = 4S³

S = 0.0118M

Hence, the Molar solubility S is 0.0118M.

Learn more about Molar solubility here, brainly.com/question/16243859

#SPJ4

6 0
2 years ago
A tank with volume of 2.4 cu ft is filled with Methane to a pressure of 1500 psia at 104 degrees F. Determine the molecular weig
Soloha48 [4]

Explanation:

It is known that equation for ideal gas is as follows.

               PV = nRT

The given data is as follows.

     Pressure, P = 1500 psia,     Temperature, T = 104^{o}F = 104 + 460 = 564 R

     Volume, V = 2.4 cubic ft,      R = 10.73 psia ft^{3}/lb mol R

Also, we know that number of moles is equal to mass divided by molar mass of the gas.

                n = \frac{mass}{\text{molar mass}}

            m = n \times W

                = 0.594 \times 16.04

                = 9.54 lb

Hence, molecular weight of the gas is 9.54 lb.

  • We will calculate the density as follows.

                d = \frac{PM}{RT}

                    = \frac{1500 \times 16.04}{10.73 \times 564}

                    = 3.975 lb/ft^{3}

  • Now, calculate the specific gravity of the gas as follows.

  Specific gravity relative to air = \frac{\text{density of methane}}{\text{density of air}}

                         = \frac{3.975 lb/ft^{3}}{0.0765 lb/ft^{3}}

                         = 51.96

6 0
3 years ago
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