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Nonamiya [84]
3 years ago
9

A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. w

hat volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Chemistry
1 answer:
o-na [289]3 years ago
4 0

Answer:

V2 = 17371.43ml

Explanation:

We use Boyles laws

since temperature is constant

P1V1=P2V2

760 x 400 = 17.5 x V2

304000 = 17.5 x V2

V2 = 304000/17.5

V2 = 17371.43ml

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zhannawk [14.2K]
You’re right its A -2
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statement 1,mass is conserved durning the reaction. statement 2 sum of mass and energy is conserved durning the reaction. which
maksim [4K]

Answer:

Explanation:

None of the statement is true for both chemical and nuclear reactions. In chemical reactions, mass is always conserved and the type of atoms are also conserved.

6 0
3 years ago
Identify the combustion reaction. identify the combustion reaction. c(s) + o2(g) → co2(g) 2 h2(g) + o2(g) → 2 h2o(g) c3h8(g) + 5
tester [92]
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7 0
3 years ago
An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g
vodomira [7]

Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity= \frac{33.8 g solute}{100 g solution} x \frac{1.15 g solution}{1 ml} x \frac{1 mol solute}{145.6 g solute} x \frac{1000 ml}{1 L}

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x \frac{1 mol solute}{145.6 g} = 0.232 mol

mol H20= 66.2 g H₂O x \frac{1 mol H2O}{18 g}

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute=\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}

Xsolute= 0.059

4 0
3 years ago
Write the net ionic equation for ammonium carbonate and lead(II) nitrate are combined. (Use the solubility rules provided in the
sashaice [31]

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of ammonium carbonate and lead (II) nitrate is given as:

(NH_4)_2CO_3(aq.)+Pb(NO_3)_2(aq.)\rightarrow 2NH_4NO_3(aq.)+PbCO_3(s)

Ionic form of the above equation follows:

2NH_4^{+}(aq.)+CO_3^{2-}(aq.)+Pb^{2+}(aq.)+2NO_3^{-}(aq.)\rightarrow PbCO_3(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+CO_3^{2-}(aq.)\rightarrow PbCO_3(s)

Hence, the net ionic equation is written above.

8 0
3 years ago
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