¿por qué la conducción ocurre más lentamente en los gases que en los sólidos?

13) The velocity of an object dropped in air will continue to increase

melamori03 [73]

Answer:

Because due to the presence air resistance

7 0

3 years ago

If 1.8 L of water is added to 2.5L of a 7.0 Molarity

liq [111]

Answer:

4.1 mol·L⁻¹

Explanation:

You are diluting the solution with water, so you can use the dilution formula

c₁V₁ = c₂V₂ Divide each side by V₂

c₂ = c₁ × V₁/V₂

Data:

c₁ = 7.0 mol·L⁻¹; V₁ = 2.5 L

Water added = 1.5 L

Calculations:

(a) New volume

V₂ = 2.5 L + 1.8 L

= 4.3 L

(b) New concentration

c₂ = 7.0 × 2.5/4.3

= 7.0 × 0.581

= 4.1 mol·L⁻¹

6 0

3 years ago

Consider the following reaction of the commercial production of SO3 using sulfur dioxide and oxygen: SO2(g) + O2(g) → SO3(g)

kipiarov [429]

Answer:

$m_{SO_3}=18.93gSO_3$

$V_{SO_3}=5.3LSO_3$

Explanation:

Hello,

STP conditions are P=1 atm and T=273.15 K, thus, the reacting moles are:

$n_{SO_2}=\frac{5.3L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2366molSO_2\\n_{O_2}=\frac{4.7L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2098molO_2$

Now, the balanced chemical reaction turns out into:

$2SO_2(g) + O_2(g) --> 2SO_3(g)$

Thus, the exact moles of oxygen that completely react with 0.2366 moles of sulfur dioxide are (limiting reagent identification):

$0.2366molSO_2*\frac{1molO_2}{2molSO_2}=0.1182molO_2$

Since 0.2098 moles of oxygen are available, we stipulate the oxygen is in excess and the sulfur dioxide is the limiting reagent. In such a way, the yielded grams of sulfur trioxide turn out into:

$m_{SO_3}=0.2366molSO_2*\frac{2molSO_3}{2molSO_2}*\frac{80gSO_3}{1molSO_3} \\m_{SO_3}=18.93gSO_3$

By using the ideal gas equation, one computes the volume as:

$V_{SO_3}=\frac{mRT}{MP}=\frac{18.93g*0.082\frac{atm*L}{mol*K} *273.15K}{80g/mol*1atm}\\V_{SO_3}=5.3LSO_3$

It has sense for volume since the mole ratio is 2/2 between sulfur dioxide and sulfur trioxide.

Best regards.

6 0

3 years ago

The density of o2 gas at 16 degrees Celsius and 1.27atm is?

velikii [3]

Answer:

The density of O₂ gas is 1.71 $\frac{g}{L}$

Explanation:

Density is a quantity that allows you to measure the amount of mass in a given volume of a substance. So density is defined as the quotient between the mass of a body and the volume it occupies:

$density=\frac{mass}{volume}$

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

So, you can get:

$\frac{n}{V} =\frac{P}{R*T}$