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2 years ago

On the reaction below, label the BSA, BSB, CA, and CB. CH3COOH + H2O → CH3COO– + H3O+

Chemistry

1 answer:

Ostrovityanka [42]2 years ago

7 0

Answer:

Acid(BSA) = CH₃COOH

Base (BSB) = H₂O

Conjugate base (CB) = CH₃COO⁻

Conjugate acid (CA) = H₃O⁺

Explanation:

Equation of reaction;

CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺

Hello,

From my understanding of the question, we are required to identify the

1) Acid

2) Base

3) conjugate acid

4) conjugate base in the reaction

Acid (BSA) = CH₃COOH

Base (BSB) = H₂O

CA = conjugate acid = H₃O⁺

CB = conjugate base = CH₃COO⁻

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Question 11 (3 points)

Ugo [173]

Answer:

The outer core of the earth is liquid.

Explanation:

Seismic waves travel through the Earth at different speeds and are reflected and refracted diffracted differently as they encounter layers of different densities.

By analyzing these waves, they have concluded that there is an inner core of solid iron surrounded by a liquid outer core consisting of nickel and iron.

A is wrong. The Earth consists of several layers

C is wrong. Seafloor spreading is caused by the movement of tectonic plates away from each other.

D is wrong. The mantle is mostly solid rock.

4 0

3 years ago

Please help, with step by step work

natali 33 [55]

$\qquad$ ☀️ $\pink{\bf{ {Answer = \: \: 85.57g }}}$

Molar mass of $\bf Cu_2O$

$\qquad$ $\twoheadrightarrow\sf 63.546 \times 2 +16$

$\qquad$ $\pink{\twoheadrightarrow\bf 143.092 g}$

<u>As we know</u>–

1 mol = $\bf 6.02×10^{23}$ formula units

1 mol $\bf Cu_2O$ = 143.092 g = $\bf 6.02×10^{23}$ formula units

Henceforth –

$\bf 3.60×10^{23}$ formula units $\bf Cu_2O$ –

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×10^{23 }}{6.02×10^{23}}$

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23 }}}{6.02×\cancel{10^{23}}}$

$\qquad$ $\pink{:\implies\bf 85.57 g}$

5 0

2 years ago

Anyone know the answer to number 1?

Andrew [12]

The rate at which the substance dissolves in water.

5 0

3 years ago

Read the excerpt from Levitt and Dubner’s Freakonomics.

VARVARA [1.3K]

Answer:

The answer is D

Explanation:

5 0

3 years ago

Read 2 more answers

An aqueous solution of a soluble compound (a nonelectrolyte) is prepared by dissolving 33.2 g of the compound in sufficient wate

stiv31 [10]

Answer:

2710.2g/mol

Explanation:

Step 1:

Data obtained from the question. This include the following:

van 't Hoff factor (i) = 1 (since the compound is non-electrolyte)

Mass of compound = 33.2g

Volume = 250mL

Osmotic pressure (Π) = 1.2 atm

Temperature (T) = 25ºC = 25ºC + 273 = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Molar mass of compound =.?

Step 2:

Determination of the molarity of the compound.

The molarity, M of the compound can be obtained as follow:

Π = iMRT

1.2 = 1 x M x 0.0821 x 298

Divide both side by 0.0821 x 298

M = 1.2 / (0.0821 x 298)

M = 0.049mol/L

Step 3:

Determination of the number of mole of compound in the solution. This can be obtain as follow:

Molarity = 0.049mol/L

Volume = 250mL = 250/1000 = 0.25L

Mole of compound =..?

Molarity = mole /Volume

0.049 = mole / 0.25

Cross multiply

Mole = 0.049 x 0.25

Mole of compound = 0.01225 mole.

Step 4:

Determination of the molar mass of the compound. This is illustrated below:

Mole of the compound = 0.01225 mole.

Mass of the compound = 33.2g

Molar mass of the compound =.?

Mole = Mass /Molar Mass

0.01225 = 33.2/Molar Mass

Cross multiply

0.01225 x molar mass = 33.2

Divide both side by 0.01225

Molar mass = 33.2/0.01225

Molar mass of the compound = 2710.2g/mol

Therefore, the molar mass of the compound is 2710.2g/mol

6 0

2 years ago