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morpeh [17]
3 years ago
7

Can i just have the answers

Chemistry
1 answer:
JulijaS [17]3 years ago
3 0

https://www.teacherspayteachers.com/Product/Counting-Atoms-Worksheet-Editable-190185


id put em on here but i cant read them to good. Hope it helps tho!

fyi if u have other worksheets u need help on if u loo up the worksheet name and answer key itll prolly pop up.

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If the volume of an irregular sound is determined by the water displacement technique, its volume will be equal to
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25 mL
i could be wrong but hope it helps
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2 years ago
If 17.4 mL of 0.800 M HCl solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentra
Verizon [17]
NH3  +HCl  ---->  NH4Cl

moles   of  HCl  used =    (0.8  x  17.4) /1000=  0.0139 moles
by  use   of reacting  ratio  between  HCl  to   NH4Cl    which is 1:1    therefore  the  moles   of  NH4Cl  is  also  =  0.0139 moles
molar   concentration  =  moles  /volume  in  liters 

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8 0
3 years ago
Read 2 more answers
A 0.5 mol sample of He(g) and a 0.5 mol sample of Ne(g) are placed separately in two 10.0 L rigid containers at 25°C. Each conta
KiRa [710]

Answer:

Helum (He)g will escape faster

Explanation:

the phenomemenon can be explained by the Graham's law of diffusion.

Graham's law of difussion states that the rate of difussion is inversely proportional to the square root of the molecular mass,which means the gas with lower molecular mass will escape faster.

Helium gas has a molecular mass of 4 while Neon has a molecular mass of 10.

rate of diffusion of He/rate of difussion of Ne=√4/10=√0.4=0.63

It means He(g) will move 0.63 times faster than Ne(g) under the same condition

5 0
3 years ago
An instrument that is used to measure air pressure is a ?
Zolol [24]
A barometer measures atmospheric pressure.
8 0
3 years ago
Which of the following acids should be used to prepare a buffer with a pH of 4.5?A. HOC6H4OCOOH, Ka = 1.0 x 10^-3B. C6H4(COOH)2,
ioda

Answer:

C. CH3COOH, Ka = 1.8 E-5

Explanation:

analyzing the pKa of the given acids:

∴ pKa = - Log Ka

A. pKa = - Log (1.0 E-3 ) = 3

B. pKa = - Log (2.9 E-4) = 3.54

C. pKa = - Log (1.8 E-5) = 4.745

D. pKa = - Log (4.0 E-6) = 5.397

E. pKa = - Log (2.3 E-9) = 8.638

We choose the (C) acid since its pKa close to the expected pH.

⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):

  • pH = pKa + Log ([CH3COO-]/[CH3COOH])

∴ pH = 4.5

∴ pKa = 4.745

⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])

⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])

⇒ 0.5692 = [CH3COO-]/[CH3COOH]

∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]

⇒ 1.8 E-5 = [H3O+](0.5692)

⇒ [H3O+] = 3.1623 E-5 M

⇒ pH = - Log ( 3.1623 E-5 ) = 4.5

6 0
3 years ago
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