Iv'e got a long one

What is the volume of 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)?

uysha [10]

Given:
 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)
Required:
volume of CL2
Solution:
Use the ideal gas law
PV = nRT
V = nRT/P
V = (2.1 moles Cl2) (0.08203 L - atm / mol - K) (273K) / (1 atm)
V = 47 L

5 0

3 years ago

In writing a chemical equation that produces hydrogen gas, the correct representation of hydrogen gas is

Dennis_Churaev [7]

Answer:

h2

Explanation:

5 0

2 years ago

Someone please help me sort these out for either if they are Heterogenous, solution, suspension, or colloid. Tuna casserole, gre

Sphinxa [80]

Heterogeneous-a mixture of two or more things
solution-a substance that dissolves
suspension-mixture where solid particles do not dissolve
colloid-does not settle and cannot be seperated

Heterogeneous
- tuna casserole
-chocolate chip cookies
-gelatin dessert
-green salad
Solution
-oil (?) depends
Suspension
-vinegar salad dressing
-oil (?) it could go in either catagory
Colloid
-vinegar salad dressing
-oil (?)

7 0

3 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

To calculate the $E^o_{(Cu^{2+}/Cu)}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)$

$E^o_{(Cu^{2+}/Cu)}=0.34V$

Hence, the value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

8 0

3 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$