All categories

3 years ago

The vapor pressure of ethanol at its normal boiling point would be

1 answer:

4 0

The normal boiling point of ethanol is 78.4 degrees C and, at thistemperature, the vapor pressure is 101325 Pascals (Pa) or 760manometric units

thx hope this helped bye.

You might be interested in

The following five beakers, each containing a solution of sodium chloride (NaCl, also known as table salt), were found on a lab

mixas84 [53]

Answer:

See the answers below

Explanation:

1) 100. mL of solution containing 19.5 g of NaCl (3.3M)

2) 100. mL of 3.00 M NaCl solution (3 M)

3) 150. mL of solution containing 19.5 g of NaCl (2.2 M)

4) Number 1 and 5 have the same concentration (1.5M)

MW of NaCl = 23 + 36 = 59 g

For number 3

59 g ------------------- 1 mol

19,5 g ----------------- x

x = 19.5 x 1/59 = 0.33 mol

Molarity (M) = 0.33 mol/0.150 l = 2.2 M

For number 4,

Molarity (M) = 0.33mol/0.10 l = 3.3 M

For number 5

Molarity (M) = 0.450/0.3 = 1.5 M

4 0

3 years ago

Read 2 more answers

Is this correct? Plz answer ASAP!

Angelina_Jolie [31]

YES IT'S @LL CORRECT

3 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d

worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B\rightarrow Products$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

$3.20\times 10^{-1}=k(1.50)^a(1.50)^b$ ....(1)

Expression for rate law for second observation:

$3.20\times 10^{-1}=k(1.50)^a(2.50)^b$ ....(2)

Expression for rate law for third observation:

$6.40\times 10^{-1}=k(3.00)^a(1.50)^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0$

Dividing 1 from 3, we get:

$\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[A]^1[B]^0$

$\text{Rate}=k[A]$

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

7 0

3 years ago

How many moles are in 12.9 grams of Fe?

amid [387]

Answer:

0.1677 is the answer

Explanation:

6 0

2 years ago

Read 2 more answers

What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)

blsea [12.9K]

the pH of a 0.050 M triethylamine, is 11.70

For triehtylamine, $(C_{2}H_{5})_{3}N$ , the reaction will be
$(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}$

and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value = 5.3 x 10^-4

And $kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}$

thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

=0.00516 M

Thus, pOH = 2.30
pH = 14 - pOH = 11.7

6 0

3 years ago