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meriva
3 years ago
8

What is the pH of 15 mL of .2 M HCl

Chemistry
1 answer:
fenix001 [56]3 years ago
3 0
Easy peasy! All we need to do is plug this formula into our calculator:

-log(M)

So, we'd plug in -log(.2), which is 0.7 :)
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A solid mixture consists of 47.6g of KNO3 (potassium nitrate) and 8.4g of K2SO4 (potassium sulfate). The mixture is added to 130
IgorLugansk [536]

<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.

<u>Explanation:</u>

We are given:

Mass of potassium nitrate = 47.6 g

Mass of potassium sulfate = 8.4 g

Mass of water = 130. g

Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g

This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water

Applying unitary method:

In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams

So, in 130 grams of water, the amount of potassium sulfate dissolved will be \frac{7.4}{100}\times 130=9.62g

As, the soluble amount is greater than the given amount of potassium sulfate

This means that, all of potassium sulfate will be dissolved.

Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.

7 0
3 years ago
Oxidation number of au in kaucl4
DanielleElmas [232]
KauCl4 :

K = + 1 

au = + 7

Cl = - 2

hope this helps!

3 0
3 years ago
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Hydrogen sulfide gas H2S and aqueous iron(II) chloride are formed by the reaction of aqueous hydrochloric acid HCl and solid iro
jasenka [17]

Answer:

The balanced chemical equation for this reaction is:

2HCl (aq) + FeS (s) -----> FeCl2 (aq) + H2S (g)

Explanation:

5 0
3 years ago
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C (graphite) is used as a lubricant, whereas C (diamond) is used as an abrasive. Why is this?
zysi [14]

Answer:

Carbon atoms in graphite and diamond are arranged in different ways. Hence, the two allotropes of carbon have different physical properties.

Explanation:

Both graphite and diamond are both made of only carbon atoms. However, their physical properties differ from each other. Hence, they are called allotropes. Think about how these carbon atoms are arranged in each of the allotropes.

<h3>Graphite</h3>

In graphite, each carbon atom is bonded to three other carbon atoms. These carbon atoms will be located in the same plane. A chunk of graphite can contain many of these planes.

Each carbon atom has four valence electrons. Three of these electrons will be used in the bonds. The other electron will be delocalized. These electrons would flow between the sheets of carbon atoms. That keeps the sheets separate and allow them to slide on top of each other.

<h3>Diamond</h3>

In diamond, each carbon atom is bonded to four other carbon atoms. These carbon atoms will form a tetrahedral network.  

In graphite, there's a significant separation between two adjacent sheets of carbon atoms. The force between the two sheets is rather weak. When a piece of graphite is between two objects that move over one another, the layers in the graphite would also slide over one another. Since the attraction between two adjacent sheets isn't very strong, there wouldn't be much resistance. Hence the graphite acts as a lubricant.  

In contrast, most of the carbon atoms in a piece of diamond would be connected to each other. Unlike the sheets in graphite, in a diamond there are almost no moving parts. Also, the forces between neighboring carbon atoms are very strong. When an external force acts on a chunk of diamond, the carbon atoms would barely move. Hence, the structure appears to be very rigid. That gives diamond its abrasive properties.

4 0
4 years ago
Calculate the amount of energy , in Joules, required to raise the temperature of 15.5 g of liquid water from 0.00o C to 75.0 oC.
deff fn [24]

Answer:

10043.225 J

Explanation:

We'll begin by calculating the amount of heat needed to change ice to water since water at 0°C is ice. This is illustrated below:

Mass (m) = 15.5g

Latent heat of fussion of water (L) = 334J/g

Heat (Q1) =..?

Q1 = mL

Q1 = 15.5 x 334

Q1 = 5177 J

Next, we shall calculate the amount of heat needed to raise the temperature of water from 0°C to 75°C.

This is illustrated below:

Mass = 15.5g

Initial temperature (T1) = 0°C

Final temperature (T2) = 75°C

Change in temperature (ΔT) = T2 – T1 = 75 – 0 = 75°C

Specific heat capacity (C) of water = 4.186J/g°C

Heat (Q2) =?

Q2 = MCΔT

Q2 = 15.5 x 4.186 x 75

Q2 = 4866.225 J

The overall heat energy needed is given by:

QT = Q1 + Q2

QT = 5177 + 4866.225

QT = 10043.225 J

Therefore, the amount of energy required is 10043.225 J

8 0
3 years ago
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