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3 years ago

Lead (11) oxide dissolves in both

Chemistry

1 answer:

inn [45]3 years ago

7 0

Answer:

bacic

Explanation:

it is very soluble in water

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2. A quantity of 1.922g of methanol (CH3OH) was burned in a constant-volume

Cerrena [4.2K]

Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:

5 0

2 years ago

a radio controlled airplane is moving at a velocity of 23 m/s. it accelerates to a velocity of 85 m/s in 9.3 seconds. what accel

tatuchka [14]

I will show you a method to find a velocity of anything:-

Find average velocity when acceleration is constant.

Set up an equation with position and time instead.

Find the distance between the start and end points.

Calculate the change in time.

Divide the total displacement by the total time.

Solve problems in two dimensions.

7 0

3 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

3 years ago

Which solution is the most concentrated? 2.0 mL of 10 M H2SO4, where H2SO4 has a molar mass of 98 g/mol 5.0 mL of 1.0 M PbSO4, w

marishachu [46]

Answer: the option 4) 2.0 mL of 10.5 M H₂O₂, where H₂O₂ has a molar mass of 34 g/mol.

Its concentration is 10.5 M.

Explanation:

1) The unit M means molar. It is the molarity of the solution.

Molartity is the concentration of the solution expressed as number of moles of solute per liters of solution.

The formula of molarity, M, is:

M = number of moles of solute / volume of solution in liters

2) 2.0 mL of 10 M H₂SO₄, where H₂SO₄ has a molar mass of 98 g/mol

⇒ concentration is 10 M

3) 5.0 mL of 1.0 M PbSO₄, where PbSO₄ has a molar mass of 303 g/mol

⇒ concentration = 1.0 M

4) 2.0 mL of 10.5 M H₂O₂, where H₂O₂ has a molar mass of 34 g/mol

⇒ concentration is 10.5 M

5) 100 mL of 10 M NaCl, where NaCl has a molar mass of 58 g/mol

⇒ concentration is 10 M

7 0

3 years ago

Read 2 more answers

To find the area of a rectangle, multiply the length by the width. Why is area called a derived quantity?

shusha [124]

Explanation:

Derived quantities are quantities that are calculated from two or more measurements. They include area, volume, and density. The area of a rectangular surface is calculated as its length multiplied by its width. The volume of a rectangular solid is calculated as the product of its length, width, and height.

5 0

3 years ago