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cluponka [151]
3 years ago
7

What is the best description for a solution with a hydroxide-ion concentration equal to 1 x 10-4 m?

Chemistry
1 answer:
djverab [1.8K]3 years ago
7 0

Answer:

This is a basic solution

Explanation:

To be able to provide a succint description of this solution, from the given parameters we can deduce the pH of the solution.

Given parameters:

Concentration of OH⁻ = 1 x 10⁻⁴M

Solution

          We knonw that pOH = -log₁₀[OH⁻]

                   pOH = -log₁₀ 0.0001 = - (-4) = 4

For an aqeous solution:

                  pH + pOH = 14

                  pH = 14 - 4 = 10

The pH of the solution has been derieved to be 10

This is a basic solution

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Which protists lack cilia and flagella, but can still move around?
timama [110]

Answer: amoebae !

Explanation:

3 0
4 years ago
How many grams of Na2So4 would be formed if 0.75 moles of NaOH reacted?
Nata [24]

Answer:

\boxed{\text{53 g }}

Explanation:

You don't give the reaction, but we can get by just by balancing atoms of Na.

We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

M_r:                                 142.04  

             2NaOH + … ⟶ Na₂SO₄ + …  

n/mol:      0.75

1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.

Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH

= 0.375 mol Na₂SO₄

2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.

Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄

The reaction produces \boxed{\text{53 g }} of Na₂SO₄.

4 0
3 years ago
Which of the following would you except to see in the death of a star that is less than 0.5 solar mass
Ket [755]

B. White Dwarf.

<h3>Explanation</h3>

The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.

As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:

Chandrasekhar Limit: 1.4 \;M_\odot.

A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its 0.5 \;M_\odot, which is much smaller than the Chandrasekhar Limit.

As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.

7 0
3 years ago
What are the coefficients of this chemical equation?
777dan777 [17]

Answer:

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8 0
4 years ago
Read 2 more answers
If 21.5 mol of an ideal gas is at 9.65 atm at 10.90 ∘C, what is the volume of the gas?
mamaluj [8]
<h3>Answer:</h3>

51.93 L

<h3>Explanation:</h3>

From the question we are given the following components of an ideal gas;

Number of moles = 21.5 mol

Pressure, P = 9.65 atm

Temperature, T = 10.90°C, but K= °C + 273.15

                         =284.05 k

We are required to calculate the volume of the ideal gas.

We are going to use the ideal gas equation which is given by;

PV = nRT, where P, V, T and n are the pressure, volume, temperature and moles of the ideal gas respectively. R is the ideal gas constant, 0.082057 L.atm/mol.K

To get the volume, we rearrange the formula to get;

V = nRT ÷ P

  = (21.5 × 0.082057 × 284.05 K) ÷ 9.65 atm

  = 51.93 L

Thus, the volume of the ideal gas is 51.93 L

3 0
3 years ago
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