Methanol and ethanol are said to be miscible in one another. What does this tell you?

A student placed equal volumes of honey and of water in two, identical, open dishes, and left them at room temperature for 8 hou

Ahat [919]

Answer: Honey has a much lower vapor pressure than pure water has. So, pure water evaporates at a much higher rate.

Explanation:

8 0

3 years ago

Read 2 more answers

Calculate the volume of the gas when the pressure of the gas is 1.60 atm at a temperature of 298 K. There are 160. mol of gas in

LekaFEV [45]

Answer:

2445 L

Explanation:

Given:

Pressure = 1.60 atm

Temperature = 298 K

Volume = ?

n = 160 mol

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 08206 L.atm/K.mol

Applying the equation as:

1.60 atm × V = 160 mol × 0.08206 L.atm/K.mol × 298 K

Answer to four significant digits, Volume = 2445 L

6 0

3 years ago

Find the empirical formula of a compound containing: 19.32% Ca, 34.30% Cl, and 46.38% O

Bess [88]

40×19.32/100=7.7=8×2=16Ca
35.5×34.30/100=12.1=12×2=24Cl
16×46.38/100=7.4=7×2=14O

5 0

3 years ago

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.