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3 years ago

Look at the picture of the fragmented rock.

Chemistry

1 answer:

Liono4ka [1.6K]3 years ago

7 0

Answer:

physical weathering

Explanation:

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Electron configuration of ag

stepladder [879]

Silver/Electron configuration

[Kr] 4d¹⁰ 5s¹

Atomic number: 47

Symbol: Ag

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4 0

3 years ago

A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 385.mg of oxalic acid H2

Vlada [557]

Answer:

The molarity of the sodium hydroxide solution is 0.0692 M

Explanation:

Step 1: Data given

Mass of H2C2O4 = 385 mg = 0.385 grams

volume = 250 mL = 0.250 L

Volume of NaOH = 123.7 mL = 0.1237 L

Molar mass H2C2O4 = 90.03 g/mol

Step 2: The balanced equation

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

Step 3: Calculate moles H2C2O4

Moles H2C2O4 = mass H2C2O4 / molar mass H2C2O4

Moles H2C2O4 = 0.385 grams / 90.03 g/mol

Moles H2C2O4 = 0.00428 moles

Step 4: Calculate molarity of H2C2O4

Molarity H2C2O4 = moles / volume

Molarity H2C2O4 = 0.00428 moles / 0.250 L

Molarity H2C2O4 = 0.01712 M

Step 5: Calculate molarity of NaOH

2*Ca*Va = n*Cb*Vb

⇒ with Ca = Molarity of H2C2O4 = 0.01712 M

⇒ Va = volume of H2C2O4 = 0.250 L

⇒Cb = molarity of NaOH = TO BE DETERMINED

⇒ Vb = volume of NaOH = 0.1237 L

Cb = (2*0.01712*0.250)/0.1237

Cb = 0.0691 M

The molarity of the sodium hydroxide solution is 0.0692 M

4 0

3 years ago

An engineer that designs buildings is a (n)

guapka [62]

An architectural engineer.

7 0

3 years ago

SHOW WORK

Makovka662 [10]

Answer:

1.51367e+10 inches

Explanation:

1 mile = 63360

63360 x 238900 = 15136704000

Hope this helped!

6 0

2 years ago

An equimolar mixture of acetone and ethanol is fed to an evacuated vessel and allowedto come to equilibrium at 65°C and 1.00 atm

frutty [35]

The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

xA= 0.34, yA= 0.55

ii) 0.50 = 0.55 nv + 0.34 nL

Therefore, nV = 0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

iii) ρA= 0.791 g/cm3 and, ρE = 0.789 g/cm3

Therefore, ρ = 0.790 g/cm3

Now, we have:

MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor

Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3

Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %

3 0

3 years ago