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myrzilka [38]
3 years ago
10

How many liters would I need to make a 4M KCl solution using 8 moles of KCI?

Chemistry
1 answer:
hammer [34]3 years ago
5 0
M=n/v
4=8/v
4v=8
v=2
I don’t know which units you were given
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Which of the following best describes the relation ship between oxygen and chlorine
emmasim [6.3K]

Answer:

D. oxygen atoms have twice as many protons as chlorine atoms

6 0
2 years ago
Explain why anhydrous aluminium chloride is fairly soluble in organic solvent while anhydrous magnesium chloride is insoluble​
prohojiy [21]

Aluminium chloride is covalent hence soluble in organic solvent while magnesium chloride being ionic is insoluble in organic solvent

8 0
3 years ago
How much heat must your body transfer to 500.0g of water to heat it from 25.0°C to body temperature, 37.0°C?
shtirl [24]

Answer : The heat your body transfer must be, 25.1 kJ

Explanation :

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat = ?

m = mass of water = 500.0 g

c = specific heat of water = 4.18J/g^oC

T_1 = initial temperature  = 25.0^oC

T_2 = final temperature  = 37.0^oC

Now put all the given value in the above formula, we get:

Q=500.0g\times 4.18J/g^oC\times (37.0-25.0)K

Q=25080J=25.1kJ

Therefore, the heat your body transfer must be, 25.1 kJ

3 0
3 years ago
What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i
UNO [17]

Answer:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)

The net equation is then:

Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-

Actual cathode half-equation:

Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)

Actual net reaction:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

6 0
3 years ago
1. Analysis of an unknown substance formerly used in rocket fuel reveals a composition of 93.28% nitrogen and 6.72% hydrogen by
Nitella [24]

Answer:

The formula of the compound is:

N2H2

Explanation:

Data obtained from the question:

Nitrogen (N) = 93.28%

Hydrogen (H) = 6.72%

Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:

N = 93.28%

H = 6.72%

Divide by their molar mass

N = 93.28 /14 = 6.663

H = 6.72 /1 = 6.7

Divide by the smallest

N = 6.663 / 6.663 = 1

H = 6.72 /6.663 = 1

Therefore, the empirical formula is NH.

Now, we can obtain the formula of the compound as follow:

The formula of a compound is simply a multiple of the empirical formula.

[NH]n = 30.04

[14 + 1]n = 30.04

15n = 30.04

Divide both side by 15

n = 30.04/15

n = 2

Therefore, the formula of the compound is:

[NH]n => [NH]2 => N2H2

6 0
3 years ago
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