What type of reaction is the digestion of solid copper wire by nitric acid?

This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): An analytical chemist has det

murzikaleks [220]

Answer:

0.11 mol

Explanation:

This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of oxygen in a sample of acetic acid. How many moles of hydrogen are in the sample?

Step 1: Given data

Formula of acetic acid: CH₃CO₂H

Moles of oxygen in the sample of acetic acid: 0.054 moles

Step 2: Establish the appropriate molar ratio

According to the chemical formula of acetic acid, the molar ratio of H to O is 4:2.

Step 3: Calculate the moles of atoms of hydrogen

We will use the theoretical molar ratio for acetic acid.

0.054 mol O × (4 mol H/2 mol O) = 0.11 mol H

3 0

2 years ago

A 1.00 l solution contains 3.50×10-4 m cu(no3)2 and 1.75×10-3 m ethylenediamine (en). the kf for cu(en)22+ is 1.00×1020. what is

Alik [6]

Answer: A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en). contains 0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+ Kf for Cu(en)2^2+ is 1x10^20. so 1 Cu+2 & 2 en --> Cu(en)2^2+ Kf = [Cu(en)2^2+] / [Cu+2] [en]^2 1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2 [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6) Cu+2 = 9.26 e-19 Molar since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2

4 0

2 years ago

The reaction A + B → 2 C has the rate law rate = k[A][B]³. By what factor does the rate of reaction increase when [A] remains co

lawyer [7]

Answer:

8

Explanation:

To solve this question, we just need to put the new number into the equation. If [A] remain constant then that mean [A2]= [A1]. If B doubled, then that mean [B2]= 2[B2]. To find what factor does the rate of reaction increases, we need to divide the first reaction rate with the second. The calculation will be:

rate2/rate1= k[A2][B2]³ / k[A1][B1]³

rate2/rate1= [A1][2B1]³ / [A1][B1]³

rate2/rate1= A1*8B1³ / A1*B1³

rate2/rate1= 8/1= 8

The rate of reaction will be 8 times faster.

7 0

2 years ago

Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter

Margarita [4]

Answer:

thickness = 0.29 cm

Explanation:

In order to make fake iron ball [made of gold] we should get mass of fake ball

should be equal to that of Iron ball.So for that we should calculate

volume of iron ball using diameter given ;formula is 4/3 pi r^3

given d= 6 cm;so radius r= 6/2 = 3 cm

then volume of Iron ball = 4/3 *3.14* 3^3 = 113.04 cm^3

So mass of iron ball = volume x density = 113.04 * 5.15 g/cm^3 = 582.156 g

This must be the mass of gold ball ;now calculate volume of gold ball using its

density

volume of gold ball = mass of gold ball/density of gold ball = 582.156 g/19.3 g/cm^3

= 30.1635 cm^3

Now this must be the volume of hollow sphere whose outer radius R = 3cm

and inner radius r= ??

Volume of hollow ball = 4/3pi[R3-r^3]

30.1635 cm^3 = 4/3 pi [3^3-r^3]

30.1635* 3/4*3.14 = 27 - r^3

7.2046 = 27- r^3

r^3 = 19.7954

r= 2.7051 cm

So the thick ness = outer radius- inner radius = 3 - 2.7051 = 0.2949 cm

rounding to 2 significant figures

we get thickness = 0.29 cm

8 0

2 years ago

A 7.591-9 gaseous mixture contains methane (CH4) and butane

mestny [16]

Answer:

65.71%

Explanation:

First, we can write the mass of the mixture, thus:

7.519g = X + Y (1)

Where X is the mass of methane and Y the mass of butane

Also, the reactions of combustion are:

CH₄ + 2O₂ → CO₂ + 2H₂O

2 moles of oxygen react per mole of methane

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

13/2 moles of oxygen react per mole of methane

That means, in therms of moles of oxygen we can write:

0.9050 moles = 2X/16.04 + 13/2Y/ 58.12

0.9050 = 0.12469X + 0.11184Y (2)

Where 16.04 and 58.12 are molar masses of methane and butane

That is because if X is the mass of methane:

X g Methane * (1mol / 16.04g) = Moles methane

Moles methane * (2 moles Oxygen / mole methane) = Moles oxygen

Replacing (1) in (2):

0.9050 = 0.12469X + 0.11184 (7.519 - X)

0.9050 = 0.12469X + 0.841 - 0.11184X

0.0641 = 0.01285X

X = 4.988g = Mass of methane.

And mass percent of methane is:

4.988g / 7.591g * 100

7 0

2 years ago