How would the freezing point of seawater compare with that of pure water?

A sample of trifluoroacetic acid, C2HF3O2, contains 38.9 g of oxygen. Calculate the mass of the trifluoroacetic acid sample.

qwelly [4]

Answer:

138.6 g of C₂HF₃O₂ have 38.9 g of O

Explanation:

Trifluoroacetic acid → C₂HF₃O₂

Molar mass = Mass of C . 2 + Mass of H + Mass of F . 3 + Mass of O .2

Molar mass = 12 . 2 + 1 . 1 + 19 . 3 + 16 . 2 = 114 g /mol

1 mol of C₂HF₃O₂ has:

2 moles of C

1 mol of H

3 moles of F

2 moles of O

If we state the relation in mass by g we say:

114 g of C₂HF₃O₂ have 24g of C, 1 g of H, 57 g of F and 32 g of O

Let's make a rule of three:

32 g of O are contained in 114 g of C₂HF₃O₂

38.9 g of O may be contained in (38.9 . 114) / 32 = 138.6 g

8 0

3 years ago

QUESTION 1<br> 1.What would happen if producers were removed from the food chain?

gregori [183]

It would cause the whole food chain to collapse.

explanation: primary consumers or herbivores, which feed of producers would directly die off. Higher level consumers would suffer as organisms from lower tropic levels start to die off

5 0

3 years ago

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Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH2)3N (Kb = 5.2 × 10−4

Nataliya [291]

Answer:

(a) 10.62

(b) 2.82

(c) 1.95

Explanation:

The neutralization reaction in this question is

(CH3CH2)3N + HCl ⇒ (CH3CH2)3NH⁺ + Cl⁻

The problem can be solved by calculating the number of moles of triethylamine after addition of the portions of HCl. Since it is a weak base if it is not consumed completely, that is in excess we will have a buffer of a waek base. If its consumed completely the pH will be determined by the strong acid HCl.

The pOH for a buffer of a weak base is gven by

pOH = pKb + log [(CH3CH2)3NH⁺] / [(CH3CH2)3N]

(a) 11 mL of 0.100 M HCl

mol HCl = 0.011 L x 0.100 mol/L = 0.0011 mol HCl

mol (CH3CH2)3N reacted = 0.0011 mol

mol (CH3CH2)3NH⁺ produced = 0.0011 mol

mol (CH3CH2)3N initially = 0.020 L x 0.1000 mol/L 0.0020 mol

mol (CH3CH2)3N left = 0.0020 mol - 0.0011 = 0.0009 mol

pKb = - log Kb = - log (5.2 x 10⁻⁴) = 3.284

Now we can compute pOH,

pOH = 3.284 + log ( 0.0011 / 0.0009 ) = 3.37

pH = 14 - pOH = 14 - 3.37 = 10.62

(b) 20.60 mL HCl

mol HCl = 0.0206 L x 0.100 mol/L = 0.00206

mol (CH3CH2)3N consumed = 0.0020 mol

This is so because the acid will consume completely the 0.0020 mol of the weak base we had originally present.

Now the problem circumscribes to that of calculating the pH of the unreacted HCl

Total Vol = 0.0206 L + 0.02 L = 0.0406 L

mol HCl = 0.0206 L x .100 = 0.00206 mol

mol HCl left = 0.00206 mol - 0.0020 mol = 0.00006 mol

[HCl] = 0.00006 mol / 0.0406 L = 0.0015 M

Since HCl is a strong acid ( 100 % ionization) :

pH = - log [H⁺] = - log ( 0.0015 ) = 2.82

(c) We will compute the pH in the same way we did for part (b)

mol HCl = 0.025 L x 0.100 mol/L = 0.0025 mol

mol HCl left = 0.0025 mol - 0.0020 mol = 0.0005

Total Volume = 0.020 L + 0.025 L = 0.045 L

[HCl] = 0.0005 mol / 0.045 L = 0.111

pH = - log ( 0.111) = 1.95

3 0

3 years ago

Question 4

xenn [34]

Answer:

there are 19 neutrons

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3 years ago

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At stp, a 12.0 liter sample of CH4 has the same total number of molecules as

Elden [556K]

Option 2: 12.0 L of $CO_{2} (g)$ at STP.

The standard pressure and temperature values are 1 atm and 273.15 K.

Using the ideal gas equation, number of moles of gas can be calculated which is as follows:

PV=nRT...... (1)

Here, P is pressure, V is volume, n is number of moles, R is gas constant and T is temperature.

Also, in 1 mole of any gas there are $6.023\times 10^{23}$ molecules of the gas. This is known as Avogadro's number and denoted by symbol $N_{A}$

Thus,

$N=n\times N_{A}$

Equation (1) can be rewritten as follows:

$PV=\frac{N}{N_{A}}RT$

On rearranging,

$N=\frac{P\times V\times N_{A}}{R\times T}$

Here, all the other terms are constant except volume, thus, gas with volume equal to the volume of $CH_{4}$ will have same number of molecules.

Volume of $CH_{4}$ gas and $CO_{2}$ gas is same thus, $CH_{4}$ will have same total number of molecules as $CO_{2}$ gas.

5 0

3 years ago

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