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Temka [501]
2 years ago
10

Ammonia, NH3, is a weak base with a Kb = 1.8 x 10-5. In a 0.8M solution of ammonia, which has a higher concentration: ammonia (N

H3) or its conjugate acid the ammonium ion (NH4+)?
Chemistry
2 answers:
nika2105 [10]2 years ago
7 0

Answer:

11

Explanation:

Mamont248 [21]2 years ago
6 0

Answer is: concentration ammonia is higher than concentration of ammonium ion.<span>
Chemical reaction of ammonia in water: NH</span>₃ + H₂O → NH₄⁺ + OH⁻.<span>
Kb(NH</span>₃) = 1,8·10⁻⁵.<span>
c</span>₀(NH₃) = 0,8 mol/L.<span>
c(NH</span>₄⁺) = c(OH⁻) = x.<span>
c(NH</span>₃) = 0,8 mol/L - x.<span>
Kb = c(NH</span>₄⁺) · c(OH⁻) / c(NH₃).<span>
0,000018 = x² /  0,8 mol/L - x. </span>

<span>x</span>²  + 0,000018x - 0,0000144 = 0.

<span> solve quadratic equation: x = c(NH</span>₄⁺) = 3,79·10⁻<span>³ mol/L.</span>

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We would like to make a golden standard kilogram in the shape of circular cylinder. The density of gold is 19.32 g/cm3. a) Find
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Explanation:

a)  Using the provided information about the density  of gold, the sample size, thickness, and the following  equations and comersion factors, find the area of   the gold leaf:

V=l \cdot w \cdot h=A \cdot h\\m=\rho \cdot V

Gold _{\rho}=19.32 \mathrm{g} / \mathrm{cm}^{3}

1 \mu=10^{-6} \mathrm{m}

First, find the volume of the sample and then find the area of the sample.

V=\frac{m}{\rho}=\frac{27.6 \mathrm{g}}{19.32 \mathrm{g} / \mathrm{cm}^{3}} \cdot\left(\frac{0.01 \mathrm{m}}{1 \mathrm{cm}}\right)^{3} =\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}

V=A \cdot h \rightarrow A=\frac{V}{h}=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{10^{-6} \mathrm{m}} \approx 1.429 \mathrm{m}^{2}

b.  Using the provided information from part a ), the radius of the cylinder, and the following equation for the volume of a cylinder, find the length of the fiber :

V=\pi r^{2} h \rightarrow h=\frac{V}{\pi r^{2}}

h=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{\pi \cdot\left(2.5 \times 10^{-6} \mathrm{m}\right)^{2}} \approx 72778 \mathrm{m}

8 0
2 years ago
A container of hydrogen at 172 kpa was decreased to 85.0 kpa producing a new volume of 765 mL. What was the original volume
labwork [276]

Answer:

\boxed{\text{378 mL}}

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Data:

\begin{array}{rclrcl}p_{1}& =& \text{172 kPa}\qquad & V_{1} &= & \text{?} \\p_{2}& =& \text{85.0 kPa}\qquad & V_{2} &= & \text{765 mL}\\\end{array}

Calculations:

\begin{array}{rcl}172V_{1} & =& 85.0 \times 765\\172V_{1} & = & 65 025\\V_{1} & = &\textbf{378 mL}\\\end{array}\\\text{The original volume was } \boxed{\textbf{378 mL}}

6 0
2 years ago
Which statement is true about molarity and percent by mass?
Triss [41]

Answer : The correct option is, They are different units of concentration.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of solution.

Formula of molarity :

\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution in liters}}

Mass percent : It is defined as the mass of solute present in the mass of solution.

Formula of mass percent :

\text{Mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

Hence, both are the different unit of concentration.

5 0
2 years ago
Read 2 more answers
A 8.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of
Ne4ueva [31]

Answer:

C3H6.

Explanation:

Data obtained from the question:

Mass of the compound = 8g

Mass of CO2 = 24.01g

Mass of H2O = 13.10g

Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01

=> 12/44 x 24.01 = 6.5g

Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1

=> 2x1/18 x 13.1 = 1.5g

Mass of O in the compound = Mass of compound – (mass of C + Mass of H)

=> 8 – (6.5 + 1.5) = 0

Next, we shall determine the empirical formula of the compound. This is illustrated below:

C = 6.5g

H = 1.

Divide by their molar mass

C = 6.5/12 = 0.54

H = 1.4/1 = 1.

Divide by the smallest

C = 0.54/0.54 = 1

H = 1/0.54 = 2

Therefore, the empirical formula is CH2

Finally, we shall determine the molecular formula as follow:

The molecular formula of a compound is a multiple of the empirical formula.

Molecular formula = [CH2]n

[CH2]n = 44

[12 + (2x1)]n = 44

14n = 44

Divide both side by 14

n = 44/14

n = 3

Molecular formula = [CH2]n = [CH2]3 = C3H6

Therefore, the molecular formula of the compound is C3H6

4 0
2 years ago
The nucleus of an atom is dense and positively charged. What was observed when positively charged particles were radiated onto a
ankoles [38]
<span>According to my knowledge, I feel the answer is -
Particles that struck the center of the atom were repelled.

Hope this helps!
</span>
8 0
2 years ago
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