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Vikentia [17]
3 years ago
11

As the [H] In a solution decreases, what happens to the [OH-]?

Chemistry
1 answer:
valentinak56 [21]3 years ago
5 0
C. It increases and the pH stays constant.
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A sample of 1.44 g of helium and an unweighed quantity of O₂ are mixed in a flask at room temperature. The partial pressure of h
skad [1K]

Answer:

Mass of O₂ present in the mixture = 14.7 g

Explanation:

We apply the mole fraction to solve this excersise:

Mole fractio of a gas is:

Moles of X gas / Total moles = Partial pressure of X gas / Total pressure

We determine total pressure:

42 Torr + 159 Torr = 201 Torr

Sum of mole fraction = 1

Mole fraction of O₂ = 159 Torr / 201 Torr → 0.791

1 - Mole fraction of O₂ = Mole fraction of He → 1 - 0.791 = 0.209

We convert the mass of He, in order to determine the moles

1.44 g / 4g/mol = 0.36 moles

Mole fraction He = Moles of He / Total moles

0.36 moles / Total moles = 0.209

0.209 / 0.36 moles = Total moles → 0.580 moles

Mole fraction of O₂ = 0.791 = Moles of O₂ / Total moles

0.791 . 0.580 moles = Moles of O₂ → 0.459 moles

We convert the moles of O₂ to mass → 0.459 mol . 32g / 1mol = 14.7 g

4 0
3 years ago
Cleaning PDB from the test tube after the experiment could be a long and arduous process. Describe the procedure that you will u
fenix001 [56]

The unknown solid is removed by rinsing with water while the PDB is removed by rinsing with alcohol.

<h3>What are solvents?</h3>

Solvents are substances which dissolve other substances to form solutions or mixtures.

Water is used as a solvent for inorganic substances while organic solvents such as alcohol or kerosene are used as solvents for organic substances such as PDB.

Alcohol is added to the test tube containing PDB, which then dissolves. The tube is rinsed with more alcohol until all the PDB dissolves.

Therefore, the unknown solid is removed by rinsing with water while the PDB is removed by rinsing with alcohol.

Learn more about solvents at: brainly.com/question/12665236

#SPJ2

6 0
2 years ago
How does the atomic arrangement of atoms lead to its crystal structure like was seen in the sample of bronze and gold and Timber
Margaret [11]
In mineralogy and crystallography, a crystal structure<span> is a unique </span>arrangement of atoms<span> in a </span>crystal. Atomic arrangement of atoms in this kind of structure usually very symmetrical and highly ordered. Causing the component within the structure is so strong and hard to break. <span>A </span>crystal structure<span> is composed of a unit cell, a set of </span>atoms arranged<span> in a particular way; which is periodically repeated in three dimensions on a lattice. </span>Crystal structure would be commonly found in solid compounds.

7 0
4 years ago
Superheated steam at 20 bar and 450oC flows at a rate of 200 kg/min to an adiabatic turbine, where it expands to 10 bar. The tur
sergejj [24]

Explanation:

The given data is as follows.

Pressure of steam at inlet of turbine, P_{1} = 20 bar

Temperature at inlet of turbine, T = 450^{o}C

Pressure at outlet of turbine, P_{2} = 10 bar  

Mass flow rate of steam, m = 200 kg/min

Work produced by the turbine, W_{s} = 1500 kW

Steam is heated at constant pressure to its initial temperature, i.e., temperature at outlet of heat exchanger, T_{3} = 450^{o}C.

(1)   For an adiabatic turbine, the energy balance is  as follows.

           -W_{s} = m({H}_{2} - {H}_{1})

where W_{s} = work done by the turbine

                      m = mass flow rate of steam

H_{1} and H_{2} are the specific enthalpy of steam at inlet and outlet conditions of turbine.

Obtain the specific enthalpy of steam from Properties of Superheated Steam table

At 20 bar and 450^{o}C, H_{1} = 3358 kJ/kg

           H_{2} = H_{1} - \frac{W_{s}}{m}

      H_{2} = 3358 kJ/kg - \frac{1500kJ/s \times 60 s/min}{200 kg/min}

           H_{2} = 2908 kJ/kg

For P = 10 bar, H =2875 kJ/kg for T= 200^{0}C and H = 2975 kJ/kg for T=250^{o}C. Interpolate the values.

The temperature corresponding to P = 10 bar and H_{2} = 2908 kJ/kg is T = 216.5^{o}C

Therefore, the outlet temperature is T_{2} = 216.5^{o}C.

(2)     Energy balance on the heater is  as follows.

         Q = \Delta H = m(H_{3} - H_{2})

where,        Q = heat input required by the steam

                \Delta H = specific enthalpy change

           H_{3} = specific enthalpy of steam at the outlet conditions of heat exchanger

At P = 10 bar and T_{3} = 450^{o}C,  H_{3} = 3371 kJ/kg.

          Q = \frac{200 kg/min}{60 s/min} \times (3371 - 2908)kJ/kg

[/tex]

          Q = 1543.33 kJ/s

or,    Q = 1543.33 kW

Therefore, the heat input required is Q = 1543.33 kW.

8 0
3 years ago
What is this. please respond asap. 2 and 3
kolbaska11 [484]

A solid stays the same shape. The atoms a close together and don't move that much. A liquid can form to any shape, the atoms bounce around a little bit, but not much. A gas has no shape, the atoms are few and far between.

A.) Liquid

B.) Solid

C.) Gas

5 0
3 years ago
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