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2 years ago

A 2 m3 wooden block has a mass of 10 kg. What is the density of the block?

Chemistry

1 answer:

Alex73 [517]2 years ago

6 0

Answer:

Explanation:

density is equal to mass over volume

p=m/v

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Calculate the mass of nitrogen dissolved at room temperature in an 95.0 LL home aquarium. Assume a total pressure of 1.0 atmatm

harina [27]

Answer:

86.3 g of N₂ are in the room

Explanation:

First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.

We apply the mole fraction:

Mole fraction N₂ = N₂ pressure / Total pressure

0.78 . 1 atm = 0.78 atm → N₂ pressure

Room temperature → 20°C → 20°C + 273 = 293K

Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K

(0.78 atm . 95L) /0.082 . 293K = n

3.08 moles = n

Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g

8 0

2 years ago

Which of the following is the definition of an open physical system? A. a system that can be affected by the outside environment

sergey [27]

Answer:

A. a system that can be affected by the outside environment, by an exchange of matter or energy

5 0

2 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

2 years ago

Ulfur dioxide gas can react in the presence of oxygen and vanadium(V) oxide to form sulfur trioxide. Sulfur trioxide is the only

LenaWriter [7]

According to the illustration, the vanadium (V) oxide would be a catalyst.

<h3>What are catalysts?</h3>

Catalysts are substances that are utilized in reactions that are not themselves consumed in reactions but only speed up the rate of the reactions.

Catalysts speed up the rate of reactions by lowering the activation energy of the reactants.

Sulfur dioxide reacts with oxygen to produce sulfur trioxide. The vanadium (v) oxide is not consumed in the reaction. Thus it only serves as a catalyst.

More on catalysts can be found here: brainly.com/question/12260131

#SPJ1

4 0

2 years ago

Using Gizmo ,describe the main use for each fuel Coal: Petroleum: In each case, what is the end product of burning the fossil fu

lana [24]

Answer:

See explanation

Explanation:

Coal is primarily used for electricity generation. The burning of coal leads to emission of gases such as oxides of carbon, oxides of sulphur, oxides of nitrogen and water vapour. All these go straight into the atmosphere.

Petroleum is primarily used as transportation fuels. The burning of petroleum oils mostly leads to the emission of oxides of carbon and sulphur together with water vapour into the atmosphere.

6 0

3 years ago