Ask question

Ask question

All categories

2 years ago

6

When the valves are opened and the gases are allowed to mix at a constant temperature, what is the distribution of atoms in each

container? Assume that the containers are of equal volume, and ignore the volume of the tubing connecting them.

1 answer:

jek_recluse [69]2 years ago

7 0

Answer:

Equal number of atoms of each gas in each container

Explanation:

When the valves opened, the two contaienrs become one and the gases beging to mix by diffusion. This phenomenom is produced by the differeces of concentration of a gas between two points of the container.

The gases will continue diffunding util their concentration in both containers are equal.

You might be interested in

Identify the number of moles in 369 grams of calcium hydroxide. Use the periodic table and the polyatomic ion resource.

topjm [15]

Answer : The number of moles in 369 grams of calcium hydroxide is, 4.98 moles

Explanation : Given,

Mass of calcium hydroxide = 369 g

Molar mass of calcium hydroxide = 74.093 g/mole

Formula used :

$\text{Moles of calcium hydroxide}=\frac{\text{Mass of calcium hydroxide}}{\text{Molar mass of calcium hydroxide}}$

Now put all the given values in this formula, we get the moles of calcium hydroxide.

$\text{Moles of calcium hydroxide}=\frac{369g}{74.093g/mole}=4.98mole$

Therefore, the number of moles in 369 grams of calcium hydroxide is, 4.98 moles

7 0

3 years ago

Convert 3.2 ft^2 to square inches

tangare [24]

460.8 square inches

3.2 * 144 = 460.8

6 0

3 years ago

Which best compares 1 mol of sodium chloride to 1 mol of aluminum chloride?

lesya692 [45]

<span>Mol is the unit of amount of substance. It is equal to 6.02 x 10^23 molecules. Now, One mol of Sodium chloride (NaCl) contains 6.022x 10^23 molecules of NaCl. Also, the number atoms of both Na (sodium) and Cl (chlorine) will be equal. Similatly, One mol of Aluminium Chloride (AlCl3) contains 6.022x 10^23 molecules of (AlCl3) but the ratio of Al and Cl atoms will be 1:3</span>

7 0

2 years ago

Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, a

Crank

Explanation:

a. $Pb(NO_3)_2(aq) + Na_2SO_4(aq)$ → ?

$Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)$

$Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)$

$Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)$

$Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)$

b. $NiCl_2(aq) + NH_4NO_3(aq)$ →

$NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)$

No precipitation is occuring.

c. $Fe_Cl2(aq) + Na_2S(aq)$ →

$FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)$

$FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)$

$Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)$

$Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)$

d. $MgSO_4(aq) + BaCl_2(aq)$ →

$MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2$

$MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)$

$BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)$

$Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

5 0

2 years ago

Which pair shares the same empirical formula? C2H4 and C6H6 CH2 and CH4 CH3 and C2H6 CH and C2H4

hoa [83]

Answer:

CH3 and C2H6

Explanation:

4 0

3 years ago

Read 2 more answers