Y

A sample of gas occupies 10.0 L at 240°C under a pressure of

NISA [10]

Answer: 1090°C

Explanation: According to combined gas laws

(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2

where P1 = initial pressure of gas = 80.0 kPa

V1 = initial volume of gas = 10.0 L

T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K

P2 = final pressure of gas = 107 kPa

V2 = final volume of gas = 20.0 L

T2 = final temperature of gas

Substituting the values,

(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2

T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)

T2 = 513 K × (1.3375) × (2)

T2 = 1372.275 K

T2 = (1372.275 - 273) °C

T2 = 1099 °C

8 0

3 years ago

Read 2 more answers

Which of the following measurements is equal to 0.373 cL?

earnstyle [38]

1. is the correct answer

6 0

3 years ago

Geologic evidence shows that Earth's tectonic plates can move thousands of kilometers across the Earth's surface. These movement

Sergio [31]

The answer is B.
occur gradually over millions of years.

7 0

3 years ago

If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate

NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

4 0

3 years ago

Complete the mechanism for the acid-catalyzed hydrolysis of the epoxide in alcohol by adding any missing atoms, bonds, charges,

Sauron [17]

Answer:

See explanation below

Explanation:

You are missing the structure, therefore, I will do an example with one that I found on another place to try to explain.

This acid mechanism always involves carbocations, and positive charges, never negative because we are in acidic mediums.

In the first step, the lone pairs of the oxigen from the epoxide, substract one hydrogen of the reactant.

Second step, the lone pairs of the oxygen from the reactant, do a nucleophylic attack to the carbon of the epoxide. In this case, it will do it to the most substitued carbon.

Then, in the third step by acid base equilibrium, the hydrogen from the reactant that attacked, is substracted from the molecule by a molecule of water (We are in acid medium, therefore, there is traces of water) and the final structure is formed.

Check picture for mechanism:

4 0

3 years ago