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larisa [96]
3 years ago
8

a container holds 7.4 moles of gas hydrogen gas makes up 25% of the total moles in the container if the pressure is 1.24 atm wha

t is then partial pressure hydrogen
Chemistry
1 answer:
Tema [17]3 years ago
6 0
This question is based on Dalton's Law of Partial Pressure which states that "the total pressure of a system of gas is equal to the sum of the pressure of each individual gas (partial pressure).

Now, Partial Pressure of a gas = (mole fraction) × (total pressure)

⇒ Partial Pressure of Hydrogen = \frac{1}{4}   ×   \frac{1.24}{1}
                  
                                                     =   0.31 atm

        Thus the Partial Pressure of Hydrogen in the container is 0.31 atm.
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When Wolverine’s 10-pound adamantium claws are dissolved in 100 mL of 10 M nitric acid, 10.7 grams of adamantium nitrate are rec
Mazyrski [523]

The percent yield is 71.3 %.

Explanation:

Percent yield is the measure to analyze the success percentage of any experiment .The percent yield of any experiment can be obtained by the ratio of actual or experimental value to expected or theoretical value multiplied with 100.

Percent Yield = \frac{Experimental Outcome}{Theoretical Outcome} *100

So, in the present problem, we have obtained 10.7 g of adamantium nitrate from Wolverine's 10 pound claws. So the actual value or the experimental value is the amount of adamantium nitrate obtained from Wolverine's claws.

Thus, the experimental outcome is 10.7 g. While we had expected to recover 15 g of adamantium nitrate. So the theoretical outcome is 15 g.

Percent yield = \frac{10.7}{15} * 100 = 71.3 %

Thus, the percent yield is 71.3 %.

7 0
2 years ago
Use the periodic table to identify the element indicated by each electron configuration by typing in the
EleoNora [17]

Answer:

1s22s22p6: Neon (Ne)

1s22s22p63s23p3: Phosphorous (P)

1s22s22p63s23p64s1: Potassium (K)

1s22s22p63s23p64s2(im not sure what 308 is supposed to be): Calcium (Ca)

1s22s22p63s23p64s23d104p65s24d3: there is no pure element that ends 4d3 that I know of so this can either be Zirconium(Zr) if it ends in 4d2 or Niobium (Nb) if it ends in 4d4

Explanation:

you can look at the periodic table and the trends to find the rough idea of where the electron configuration ends, there are helpful articles and images on these, i attached an image that may help. After that you can look at the atomic number to find the number of electrons for a pure element and use the electron subshell pattern thing to find the exact number

5 0
1 year ago
6. A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCI. Calculate the pH
Vadim26 [7]

Answer:

a) pH = 9.14

b) pH = 8.98

c) pH = 8.79

Explanation:

In this case we have an acid base titration. We have a weak base in this case the pyridine (C₅H₅N) and a strong acid which is the HCl.

Now, we want the know the pH of the resulting solution when we add the following volume of acid: 0, 10 and 20.

To know this, we first need to know the equivalence point of this titration. This can be known using the following expression:

M₁V₁ = M₂V₂  (1)

Using this expression, we can calculate the volume of acid required to reach the equivalence point. Doing that we have:

M₁V₁ = M₂V₂

V₁ = M₂V₂ / M₁

V₁ = 0.125 * 25 / 0.1 = 31.25 mL

This means that the acid and base will reach the equivalence point at 31.25 mL of acid added. So, the volume of added acid of before, are all below this mark, so we can expect that the pH of this solution will be higher than 7, in other words, still basic.

To know the value of pH, we need to apply the following expression:

pH = 14 - pOH  (2)

the pOH can be calculated using this expression:

pOH = -log[OH⁻]  (3)

The [OH⁻] is a value that can be calculated when the pyridine is dissociated into it's ion. However, as this is a weak acid, the pyridine will not dissociate completely in solution, instead, only a part of it will be dissociated. Now, to know this, we need the Kb value of the pyridine.

The reported Kb value of the pyridine is 1.5x10⁻⁹ so, with this value we will do an ICE chart for each case, and then, calculate the value of the pH.

<u>a) 0 mL of acid added.</u>

In this case, the titration has not begun, so the concentration of the base will not be altered. Now, with the Kb value, let's write an ICE chart to calculate the [OH⁻], the pOH and then the pH:

       C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.125                                0             0

e)        -x                                   +x           +x

c)      0.125-x                              x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.125-x --> Kb is really small, so we can assume that x will be very small too, and 0.125-x can be neglected to only 0.125, and then:

1.5x10⁻⁹ = x² / 0.125

1.5x10⁻⁹ * 0.125 = x²

x = [OH⁻] = 1.37x10⁻⁵ M

Now, we can calculate the pOH:

pOH = -log(1.37x10⁻⁵) = 4.86

Finally the pH:

pH = 14 - 4.86

<h2>pH = 9.14</h2>

<u>b) 10 mL of acid added</u>

In this case the titration has begun so the acid starts to react with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.010) = 1x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 1x10⁻³ = 2.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 2.125x10⁻³ / (0.025 + 0.010) = 0.0607 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.0607                             0             0

e)        -x                                   +x           +x

c)      0.0607-x                           x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.0607-x --> 0.0607

1.5x10⁻⁹ = x² / 0.0607

1.5x10⁻⁹ * 0.0607 = x²

x = [OH⁻] = 9.54x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(9.54x10⁻⁶) = 5.02

Finally the pH:

pH = 14 - 5.02

<h2>pH = 8.98</h2>

<u>c) 20 mL of acid added:</u>

In this case the titration it's almost reaching the equivalence point and the acid is still reacting with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.020) = 2x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 2x10⁻³ = 1.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 1.125x10⁻³ / (0.025 + 0.020) = 0.025 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.025                                0             0

e)        -x                                   +x           +x

c)      0.025-x                             x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.025-x --> 0.025

1.5x10⁻⁹ = x² / 0.025

1.5x10⁻⁹ * 0.025 = x²

x = [OH⁻] = 6.12x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(6.12x10⁻⁶) = 5.21

Finally the pH:

pH = 14 - 5.21

<h2>pH = 8.79</h2>
5 0
3 years ago
Select all of the answers that apply. what are the two main ingredients in air? oxygen carbon dioxide nitrogen helium hydrogen
RSB [31]
Carbon dioxide and nitrogen
8 0
3 years ago
Read 2 more answers
Did Chelate compounds increase the rate of reaction?
Viefleur [7K]

Answer:

Chelate, any of a class of coordination or complex compounds consisting of a central metal atom attached to a large molecule, called a ligand, in a cyclic or ring structure. An example of a chelate ring occurs in the ethylenediamine-cadmium complex:

The ethylenediamine ligand has two points of attachment to the cadmium ion, thus forming a ring; it is known as a didentate ligand. (Three ethylenediamine ligands can attach to the Cd2+ ion, each one forming a ring as depicted above.) Ligands that can attach to the same metal ion at two or more points are known as polydentate ligands. All polydentate ligands are chelating agents.

Chelates are more stable than nonchelated compounds of comparable composition, and the more extensive the chelation—that is, the larger the number of ring closures to a metal atom—the more stable the compound. This phenomenon is called the chelate effect; it is generally attributed to an increase in the thermodynamic quantity called entropy that accompanies chelation. The stability of a chelate is also related to the number of atoms in the chelate ring. In general, chelates containing five- or six-membered rings are more stable than chelates with four-, seven-, or eight-membered rings.

Explanation:

7 0
2 years ago
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