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Kipish [7]
2 years ago
8

Alice needs to calculate the average pressure inside a gas cylinder. She has access to both a data table showing the pressure ov

er time and a graph of pressure over time. Which representation is probably most useful to her if she wants a precise answer?
Chemistry
2 answers:
Stella [2.4K]2 years ago
7 0
I don’t know but I will find iut
Gwar [14]2 years ago
5 0

The answer is Data Table.

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What type of pollution in cities composed of car exhaust and industrial pollutes
Troyanec [42]
Air pollution is the answer

4 0
2 years ago
Read 2 more answers
Please help: A 0.200 M NaOH solution was used to titrate a 18.25 mL HF
algol [13]

The molar concentration of the original HF  solution : 0.342 M

Further explanation

Given

31.2 ml of 0.200 M NaOH

18.2 ml of HF

Required

The molar concentration of HF

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)

Titrant = NaOH(1)

Titrate = HF(2)

Input the value :

\tt 0.2\times 31.2\times 1=M_2\times 18.25\times 1\\\\M_2=0.342

7 0
2 years ago
n acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form
Nady [450]

Answer : The ratio of the protonated to the deprotonated form of the acid is, 100

Explanation : Given,

pK_a=8.0

pH = 6.0

To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}

Now put all the given values in this expression, we get:

6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}

\frac{[Deprotonated]}{[Protonated]}=0.01  

As per question, the ratio of the protonated to the deprotonated form of the acid will be:

\frac{[Protonated]}{[Deprotonated]}=100  

Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100

5 0
2 years ago
How do I calculate the standard enthalpy change for the following reaction at 25 °C ?
ladessa [460]
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g) 

Using the standard enthalpies of formation given in the source below: 

(−601.24 kJ) + (2 x −92.30 kJ) − (−641.8 kJ) − (−285.8 kJ) = +141.76 kJ 
So: 

MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g), ΔH = +141.76 kJ
4 0
3 years ago
Question attached ! asap please
bezimeni [28]

Answer:

A. copper is highly water soluble. It will turn into 5 different hydrates as it absorbs more and more water.

b. Glycerol is easily soluble in water, due to the ability of the polyol groups to form hydrogen bonds with water molecules

c. octane is considered to be non-polar, it will not be soluble in water, since water is a polar solvent. This will happen because octane (hydrocarbons in general) contains neither ionic groups, nor polar functional groups that can interact with water molecules.

d. Nitric acid decomposes into water, nitrogen dioxide, and oxygen, forming a brownish yellow solution.

e. Barium carbonate is a white powder. It is insoluble in water and soluble in most acids

Explanation:

6 0
2 years ago
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