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Nonamiya [84]
2 years ago
12

What is the pressure of 0.60 moles of a gas if its volume is 10.0 liters at 35.0°C

Chemistry
1 answer:
eduard2 years ago
3 0

Answer: 1.52 atm

Explanation:

Given that:

Volume of gas V = 10.0L

Temperature T = 35.0°C

Convert Celsius to Kelvin

(35.0°C + 273 = 308K)

Pressure P = ?

Number of moles = 0.6 moles

Molar gas constant R is a constant with a value of 0.0821 atm L K-1 mol-1

Then, apply ideal gas equation

pV = nRT

p x 10.0L = 0.6 moles x (0.0821 atm L K-1 mol-1 x 308K)

p x 10.0L = 15.17 atm L

p = 15.17 atm L / 10.0L

p = 1.517 atm (round to the nearest hundredth as 1.52 atm)

Thus, the pressure of the gas is 1.52 atm

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An electrochemical cell is composed of pure nickel and pure iron electrodes immersed in solutions of their divalent ions at room
Andrej [43]

Answer:

0.758 V.

Explanation:

Hello!

In this case, case when we include the effect of concentration on an electrochemical cell, we need to consider the Nerst equation at 25 °C:

E=E\°-\frac{0.0591}{n} log(Q)

Whereas n stands for the number of moles of transferred electrons and Q the reaction quotient relating the concentration of the oxidized species over the concentration of the reduced species. In such a way, we can write the undergoing half-reactions in the cell, considering the iron's one is reversed because it has the most positive standard potential so it tends to reduction:

Fe^{2+}+2e^-\rightarrow Fe^0\ \ \ E\°=0.440V\\\\Ni^0\rightarrow Ni^{2+}+2e^-\ \ \ E\°=-0.250V

It means that the concentration of the oxidized species is 0.002 M (that of nickel), that of the reduced species is 0.40 M and there are two moles of transferred electrons; therefore, the generated potential turns out:

E=(0.440V+0.250V)-\frac{0.0591}{2} log(\frac{0.002M}{0.40M} )\\\\E=0.758V

Beat regards!

8 0
3 years ago
Which of the following electron configurations gives the correct arrangement of the four valence electrons of the carbon atom in
Juliette [100K]

Answer:

D) 2s^12p^3

Explanation:

Carbon.

The electronic configuration is -  

1s^22s^22p^2

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, Carbon has 2 singly occupied orbitals.

But in methane, CH_4 it forms 4 bonds. So, 1 electron each from 2s orbital jumps to the next orbital in the p subshell.

Thus, the configuration is:-

1s^22s^22p^2

Thus, the valence electron configuration is:-

2s^12p^3

7 0
2 years ago
Name the two properties of elections that indicate a wave-like nature.
tangare [24]
<span>The wave nature of an electron is indicated by its motion and the diffraction and interference of electrons in a beam.</span>
4 0
3 years ago
Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral
Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

Mass of aluminum hydroxide = 350 mg =  0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = \frac{0.350 g}{78 g/mol}=0.004487 mol

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

\frac{3}{1}\times 0.004487 mol=0.01346 mol of HCl.

Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O

Mass of magnesium hydroxide = 250 mg =  0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = \frac{0.250 g}{58 g/mol}=0.004310 mol

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310  mole of magnesium hydroxide will be neutralize by :

\frac{2}{1}\times 0.004310 mol=0.008621 mol of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}

V=\frac{0.02208 mol}{0.143 M}=0.1544 L

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2

Mass of calcium carbonate = 970mg =  0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = \frac{0.970 g}{100 g/mol}=0.00970 mol

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

\frac{2}{1}\times 0.00970 mol=0.0194 mol of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}

V=\frac{0.0194 mol}{0.143 M}=0.1357 L

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0
3 years ago
Increasing the volume of a sealed container will cause the gas particles within the container to.
Stella [2.4K]

Answer:

collide less often and with less force

3 0
1 year ago
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