All categories

3 years ago

Why do gases diffuse more quickly than liquids?

Chemistry

1 answer:

Alborosie3 years ago

5 0

Answer:

D)Gas particles move rapidly and have space between them.

Explanation:

Matter exists in three states namely: solids, liquids and gases. The particles contained in these three states are different from one another. In the gaseous state, the particles are FAR APART from one another i.e. space exists and they move at a very fast rate in contrast to the particles of a liquid, which have less space and move slower.

This rapid movement of gas particles within a less restricted space accounts for the reason why gaseous substances DIFFUSE more quickly than liquids.

You might be interested in

What is the mass of 61.9 L of oxygen gas collected at STP?

Tcecarenko [31]

Answer:

D. 44.2 g O₂

General Formulas and Concepts:
Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

Dimensional Analysis
Mole Ratio

Explanation:

Step 1: Define

Identify given.

61.9 L O₂ at STP

Step 2: Convert

We know that the oxygen gas is at STP. Therefore, we can set up and solve for how many moles of O₂ is present:

$\displaystyle 61.9 \ \text{L} \ \text{O}_2 \bigg( \frac{1 \ \text{mol} \ \text{O}_2}{22.4 \ \text{L} \ \text{O}_2} \bigg) = 2.76339 \ \text{mol} \ \text{O}_2$

Recall the Periodic Table (Refer to attachments). Oxygen's atomic mass is roughly 16.00 grams per mole (g/mol). We can use a mole ratio to convert from moles to grams:

$\displaystyle 2.76339 \ \text{mol} \ \text{O}_2 \bigg( \frac{16.00 \ \text{g} \ \text{O}_2}{1 \ \text{mol} \ \text{O}_2} \bigg) = 44.2143 \ \text{g} \ \text{O}_2$

Now we deal with sig figs. From the original problem, we are given 3 significant figures. Round your answer to the exact same number of sig figs:

$\displaystyle 44.2143 \ \text{g} \ \text{O} \approx \boxed{ 44.2 \ \text{g} \ \text{O}_2 }$

∴ our answer is letter choice D.

---

Topic: AP Chemistry

Unit: Stoichiometry

6 0

2 years ago

It is expected that a chemical reaction will occur when copper metal is combined with aqueous zinc sulfate. Explain why there wi

REY [17]

Answer:

$E_{cell}$ = +ve, reaction is spontaneous

$E_{cell}$ = -ve, reaction is non spontaneous

$E_{cell}$ = 0, reaction is in equilibrium

Case 1: when copper metal is combined with aqueous zinc sulfate.

$Cu+ZnSO_4\rightarrow Zn+CuSO_4$

Here copper is undergoing oxidation ad thus acts as anode and zinc is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential = $E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Zn^{2+}/Zn]}- E^0_{[Cu^{2+}/Cu]}$

$E^0=-0.76-(+0.34)=-1.10V$

Thus as $E_{cell}$ is negative , the reaction is non spontaneous.

Case 2: when zinc metal and aqueous copper sulfate solution are combined.

$Zn+CuSO_4\rightarrow Cu+ZnSO_4$

Here zinc is undergoing oxidation ad thus acts as anode and copper is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential = $E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Zn^{2+}/Zn]}$

$E^0=+0.34-(-0.76)=+1.10V$

Thus as $E_{cell}$ is positive , the reaction is spontaneous.

3 0

4 years ago

CO2 + H2O -------> H2CO3 balance and classify this equation

borishaifa [10]

CO2 + H2O = H2CO3

Your equation is already balanced, and the reaction type is: synthesis.

6 0

4 years ago

Read 2 more answers

How many mL of 0.563 M HNO3 are needed to dissolve 7.83 g of BaCoz? 2HNO3(aq) + BaCO3(s) — Ba(NO3)2(aq) + H2O(1) + CO2(8) mL

Lorico [155]

Answer:

Volume of HNO₃ required = 140 mL

Explanation:

Given data:

Molarity of HNO₃ = 0.563 M

mass of BaCO₃ = 7.83 g

Volume of HNO₃ = ?

Solution:

First of all we will write the balance chemical equation

2HNO₃ + BaCO₃ → Ba(NO₃)₂ + H₂O + CO₂

Number of moles of BaCO₃ = mass / molar mass

Number of moles of BaCO₃ = 7.83 g / 197.34 g/mol

Number of moles of BaCO₃ = 0.04 mol

Now we compare the moles of BaCO₃ and HNO₃ .

BaCO₃ : HNO₃

1 : 2

0.04 : 2×0.04 = 0.08 mol

Volume of HNO₃ required = number of moles / Molarity

Volume of HNO₃ required = 0.08 mol / 0.563 mol/L

Volume of HNO₃ required = 0.14 L

0.14 × 1000 = 140 mL

8 0

3 years ago

If 150 G of war started burning 300 G of where as was left behind how many grams of CO2 gas is released

denis23 [38]

Answer:

geruow0irghvn3p0unhie0ghik

Explanation:

5 0

3 years ago