A ______ is a hole where the steam looks a lot like a smoke

Why is gold usually found in its pure form

attashe74 [19]

gold is usually found in pure form because it is not reacting with other chemicals naturally.

5 0

3 years ago

What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso

Anarel [89]

The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

$pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247$

The pH of the buffer can be known as

$pH = pK_{a} + log[\frac{[A-]}{[HA]}}]$

The concentration of $[A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\$

Similarly, the concentration of [HA] = $\frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404$

Then the pH of the buffer will be

pH = 6.247 + log [ 0.304/0.404]

$pH = 6.247 + log 0.304 - log 0.404=6.247-0.517+0.3936=6.1236$

So, the pH of the buffer is 6.1236.

5 0

3 years ago

Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for

Katyanochek1 [597]

Answer:

See explanation below

Explanation:

There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.

However we do not have a pH value here.

The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.

Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:

LiOH ---> Strong. If you try to dissociate :

LiOH ------> Li⁺ + OH⁻ The Li⁺ is a weak conjugate acid.

HF -----> Weak

HF --------> H⁺ + F⁻ The Fluorine is a relatively strong conjugate base.

HCl -----> Strong

This is actually one of the strongest acid.

NH₃ ------> Weak

Now writting the Ka and Kb expressions:

Ka = [H⁺] [F⁻] / [HF]

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Finally, to calculate the [OH⁻] we need to use the following expression:

Kw = [H⁻] [OH⁻]

Solving for [OH⁻] we have:

[OH⁻] = Kw / [H⁺]

Remember that the value of Kw is 1x10⁻¹⁴. So replacing:

[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶

[OH⁻] = 1.43x10⁻⁹ M

And now, multiplying by 10¹⁰ we have:

[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰

Hope this helps

4 0

2 years ago

Thinner in the center that at the edges; virtual image smaller.

ipn [44]

<h3><u>Answer</u>;</h3>

Concave Lenses

<h3><u>Explanation</u>;</h3>

A concave lens is thin in the middle and thick at the edges, such that it seems to cave inwards. It spreads light rays apart producing an image smaller than the actual object.
<em><u>Images formed by a concave lens are virtual, upright, reduced in size and located on the same side of the lens as the object. Diverging lenses or concave lens always produce images that share these characteristics. The location of the object does not affect the characteristics of the image. </u></em>

3 0

3 years ago

Read 2 more answers

What is the freezing point of a solution of 498mL of water (solute) dissolved in 2.50 L of ethanol (solvent), C2H5OH? The densit

jok3333 [9.3K]

Answer:

Freezing T° of solution is -142.4°C

Explanation:

This excersise is about colligative properties, in this case freezing point depression,

ΔT = Kf . m . i

Where ΔT = Freezing T° of solvent - Freezing T° of solution

Kf = Cryoscopic constant

m = mol/kg (molality)

i = Number of ions dissolved.

Water is not ionic, so i = 1

Let's find out m.

We determine mass of water, by density

498ml . 1 g/mL = 498 g

We convert the mass of water to moles → 498 g . 1mol/18g = 27.6 moles

We determine mass of solvent by density

2500 mL . 0.789 g/mL = 1972.5 g

Notice, we had to convert L to mL to cancel units.

1 cm³ = 1 mL

We convert the mass from g to kg → 1972.5 g . 1kg /1000g = 1.97kg

We determine m = mol/kg → 27.6mol / 1.97kg = 13.9 m

Kf for ethanol is: 1.99 °C/m

Freezing T° for ethanol is: -114.6°C

We replace at formula: - 114.6°C - Freezing T° solution = 1.99 °C/m . 13.9 m . 1

- 114.6°C - Freezing T° solution = 27.8 °C

- Freezing T° solution = 27.8°C + 114.6°C

Freezing T° Solution = - 142.4 °C

7 0

2 years ago

A ______ is a hole where the steam looks a lot like a smoke​

A ______ is a hole where the steam looks a lot like a smoke