What is the name of GaF3?

18 g of argon occupy 750 ml at a particular temperature and pressure. How many grams of methane would occupy the same volume at

frozen [14]

Answer:

7.21 grams is the mass of methane

Explanation:

We may use the Ideal Gases Equation to solve this:

P. V = n. R. T

Let's determine the moles of Ar

18 g . 1 mol/ 39.9 g = 0.451 mol

In both situations, volume, temperature and pressure are the same so the moles of methane will also be the same as Argon's.

Let's convert the moles to mass of CH4.

0.451 mol . 16g/1mol = 7.21 grams

5 0

3 years ago

Consider the electrolysis of molten barium chloride, BaCl2. (a) Write the half-reactions. (b) How many grams of barium metal can

Zielflug [23.3K]

Answer: a) Cathode(-): $Ba^+^2+2e^-\rightarrow Ba$

Anode(+): $2Cl^-\rightarrow Cl_2+2e^-$

b) 0.640 grams of Ba will be deposited.

Explanation: a) The problem is based on Faraday law of electrolysis. Molten barium chloride has $Ba^+^2$ ion and $Cl^-$ ion. Barium ion is reduced and chloride is oxidized. Reduction takes place at cathode and oxidation at anode. So, the half reactions will be:

Cathode(-): $Ba^+^2+2e^-\rightarrow Ba$

Anode(+): $2Cl^-\rightarrow Cl_2+2e^-$

b) The question asks, how many grams of barium metal can be produced by supplying 0.50 ampere for 30 minutes.

From the Cathode half-reaction, 1 mole of Ba is deposited by 2 moles of electrons and we know that 1 mole of electron carries one Faraday that is 96485 Coulomb.

Coulombs for 2 moles of electrons will be = 2*96485 C = 192970 C

So, we can say that, 192970 C will deposit 1 mole of Ba metal.

Total available coulombs can be calculated using the formula:

q=i*t

where, q is electric charge in coulomb, i is current in ampere and t is time in seconds.

$q=q=0.50A*30min(\frac{60sec}{1min})$

q = 900 C (note: 1 C = 1 A*sec)

Let's calculate how many moles of Ba will get deposited by 900 C.

$900C(\frac{1molBa}{192970C})$

= 0.00466 mole Ba

Convert the moles of Ba to grams and for this we multiply by molar mass of Ba which is 137.33 gram per mol.

$0.00466molBa(\frac{137.33g}{1mol})$

= 0.640 g Ba

So, 0.50 A for 30 minutes will deposit 0.640 grams of Ba metal.

6 0

3 years ago

Use your periodic table and calculator as needed for the following question.

sveticcg [70]

Hope you find this answer I need points

7 0

3 years ago

A waste treatment pond is 50 m long and 25 m wide, and has an average depth of 2 m. The density of the waste is 75.3lbm/ft^3 .Ca

umka21 [38]

Answer:

$W = 6.65 \cdot 10^{6} lbf$

Explanation:

To find the weight (W) of the pond contents first we need to use the following equation:

$W = m\cdot g$ (1)

Where m the mass and g is the gravity

Also, we have that the mass is:

$m = \rho*V$ (2)

Where ρ is the density and V the volume

We cand calculate the volume as follows:

$V = L*w*d$ (3)

Where L is the length, w is the wide and d is the depth

By entering equation (2) and (3) into (1) we have:

$W = \rho*L*w*d*g$

$W = 75.3 lbm/ft^{3}*50 m*25 m*2 m*9.81 m/s^{2}$

$W = 75.3 lbm/ft^{3}*\frac{(1 ft)^{3}}{(0.3048 m)^{3}}*\frac{0.454 kg}{1 lbm}*50 m*25 m*2 m*9.81 m/s^{2} = 2.96 \cdot 10^{7} N}*\frac{0.2248 lbf}{1 N} = 6.65\cdot 10^{6} lbf$

Therefore, the weight of the pond is 6.65x10⁶ lbf.

I hope it helps you!

6 0

3 years ago

What makes an element more metallic than another?

Valentin [98]

Group names in the periodic table give clues about the metallic properties of the elements.
Metallic elements are found on the left side of the periodic table. A simple conception of metals describes them as a lattice of positive ions immersed in a sea of electrons.

8 0

3 years ago