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serious [3.7K]
2 years ago
12

Write any three acids which are used in our daily life. Also, write an application of each​

Chemistry
1 answer:
lidiya [134]2 years ago
3 0
Sodium fluoride- to brush teeth
Citric acid- orange juice for breakfast
Sodium hydroxide- cleaning agent
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Determine the velocity of a 55-kg skier whose kinetic energy is 8900 J
goldfiish [28.3K]

Answer:

Kinetic energy = 1/2mv^2

8900 j =1/2*55*v^2

v^2=8900*2/55

v^2=323.6

v=17.98

Explanation:

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Will mark brainliest
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Answer:

4. Principal and Azimuthal (subsidiary) quantum number

5.Principal, Azimuthal (subsidiary), and magnetic quantum number

6. 10 electrons

7. 32 electrons

8. 36 electrons

Explanation:

4. Principal and Azimuthal (subsidiary) quantum number because in 4d, 4 represent principal quantum number and d- represents azimuthal quantum number (having l- value as 3)

5.Principal, Azimuthal (subsidiary), and magnetic quantum number are the first three because 2 stands for principal, s-for azimuthal (l=0) and magnetic quantum number for s- orbital= 0

6. 10 electrons, because for sublevel with l= 3, is a d-sub-level, and d- can take 10-electrons

7. 32 electrons, using the relationship 2×n^2 for the maximum number of electrons in a shell,

,n= 4 , hence 2×4^2= 32

8. 36 electrons, because n=4 and n= 3 can have the maximum configuration of [Ar]4s^2 3d^10 4p^6

This will sum up to 36- electrons, since Argon has 18 -electrons.

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Answer:

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baka po pero pwd din ang B pero ok ang C

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Calculate the concentration of formate in a 100mm solution of formic acid at ph 4.15. The pka for formic acid is 3.75
MissTica

The molarity of formic acid is 100 mM or 100\times 10^{-3}M. The dissociation reaction of formic acid is as follows:

HCOOH\leftrightharpoons HCOO^{-}+H^{+}

The expression for dissociation constant of the reaction will be:

K_{a}=\frac{[HCOO^{-}][H^{+}]}{[HCOOH]}

Rearranging,

[HCOO^{-}]=\frac{K_{a}[HCOOH]}{[H^{+}]}

Here, pH of solution is 4.15 thus, concentration of hydrogen ion will be:

[H^{+}]=10^{-pH}=10^{-4.15}=7.08\times 10^{-5}M

Similarly, pK_{a}=3.75 thus,

[K_{a}=10^{-pK_{a}}=10^{-3.75}=1.78\times 10^{-4}M

Putting the values,

[HCOO^{-}]=\frac{(1.78\times 10^{-4}M)(100\times 10^{-3}M)}{(7.08\times 10^{-5}M}=0.2511 M

Therefore, the concentration of formate will be 0.2511 M.

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