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worty [1.4K]
2 years ago
10

What is an example in which a scientists might calculate the mean of a data set

Chemistry
1 answer:
gayaneshka [121]2 years ago
3 0
An example is just when they have a a group o numbers, for any reason, and they want to find the average of then all

an example being um idk...numbers resulting from an experiment trying to find different densitys or something and they wanted to find the average density
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A 25.0-ml sample of hno3 solution requires 35.7 ml of 0.108 m na2co3 to completely react with all of the hno3 in the solution. w
Rasek [7]
If we write the equation of the reaction that will take place, it is:
2HNO₃ + Na₂CO₃ → 2NaNO₃ + H₂CO3

The molar ratio of 2HNO₃ : Na₂CO₃ = 1 : 2
Therefore, we can set up the equation:
M₁V₁ = 2M₂V₂
Where the left side of the equation has the molarity and volume of HNO₃ and the right side has the molarity and concentration of Na₂CO₃. Substituting:
M₁ = (2 x 0.108 x 35.7) / 25
M₁ = 0.308 M
5 0
2 years ago
Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2
maxonik [38]

Answer:

Equilibrium constant of the given reaction is 1.3\times 10^{-29}

Explanation:

NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr....(K_{p})_{1}=5.3

2NO\rightleftharpoons N_{2}+O_{2}....(K_{p})_{2}=2.1\times 10^{30}

The given reaction can be written as summation of the following reaction-

2NO+Br_{2}\rightleftharpoons 2NOBr

N_{2}+O_{2}\rightleftharpoons 2NO

......................................................................................

N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr

Equilibrium constant of this reaction is given as-

\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}

=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})

=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}

=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}

7 0
2 years ago
Read 2 more answers
How many acidic protons are there in 0.6137 g of KHP?
andreyandreev [35.5K]

0.6137 g of KHP contains 1.086 × 10^21 acidic protons.

Number of moles of KHP = mass of KHP/molar mass of KHP

Molar mass of KHP = 204.22 g/mol

Mass of KHP = 0.6137 g

Number of moles of KHP = 0.6137 g/204.22 g/mol = 0.003 moles of KHP

Now,  1 each molecule of KHP contains 1 acidic proton.

For 0.003 moles of KHP there are; 0.003 × 1 × NA

Where NA is Avogadro's number.

So;  0.003 moles of KHP contains 0.003 × 1 × 6.02 × 10^23

= 1.086 × 10^21 acidic protons.

Learn more: brainly.com/question/16672114

7 0
2 years ago
A sample of ammonia has a mass of 82.9 g. how many molecules are in this sample?
alina1380 [7]
The chemical formula for ammonia is NH3. So first, you need to find the molar mass of ammonia (how many grams in one mole).
N=14g
H3=3g
So one mole of NH3 is 17 grams, you can divide 82.9 grams by 17 grams to find the number of molecules. The answer should be 4.876 moles (molecules) of ammonia. Hope this helps!
8 0
2 years ago
Read 2 more answers
Find the concentration of H+ ions at a pH = 11 and pH = 6. Then divide the concentration of H+ ions at a pH = 11 by the of H+ io
andrew11 [14]

Explanation:

When pH of the solution is 11.

pH=-\log[H^+]

11=-\log[H^+]

[H^+]=1\times 10^{-11} M..(1)

At pH = 11, the concentration of H^+ ions is 1\times 10^{-11} M.

When the pH of the solution is 6.

pH=-\log[H^+]'

6=-\log[H^+]'

[H^+]'=1\times 10^{-6} M..(2)

At pH = 6, the concentration of H^+ ions is 1\times 10^{-6} M.

On dividing (1) by (2).

\frac{[H^+]}{[H^+]'}=\frac{1\times 10^{-11} M}{1\times 10^{-6}}=1\times 10^{-5}

The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is 1\times 10^{-5}.

Difference between the H^+ ions at both pH:

1\times 10^{-6} M-1\times 10^{-11} M=9.99\time 10^{-7} M

This means that Hydrogen ions in a solution at pH = 7 has 9.99\time 10^{-7} M ions fewer than in a solution at a pH = 6

6 0
2 years ago
Read 2 more answers
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