Ask question

Ask question

All categories

3 years ago

15

a given sample of a gas has a volume of 3.0 L at a pressure of 4.0 atm. If the temperature remains constant and the pressure is

changed to 6.0 atm, what is the new volume?

1 answer:

TEA [102]3 years ago

6 0

According to boyle p1v1=p2v2
So 3*4=6*x so x=2

You might be interested in

Choose the correct answer !

zavuch27 [327]

Answer:

Osmium

Explanation:

The metals conduct electricity the best, in this question the metal would be osmium.

4 0

2 years ago

Help me please an thank you

agasfer [191]

Answer:

formula unit

Explanation:

3 0

2 years ago

Read 2 more answers

The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at

nalin [4]

Answer:

The given statement - The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at each other along the line between the two nuclei , is <u>True.</u>

Explanation:

The above statement is correct , because the sigma bond is produced by the head on overlapping, the orbitals should all point in the same direction.

<u>SIGMA BONDS -</u> Sigma bonds (bonds) are the strongest type of covalent chemical bond in chemistry. They're made up of atomic orbitals that collide head-on. For diatomic molecules, sigma bonding is best characterized using the language and tools of symmetry groups.

Head-on overlapping of atomic orbitals produces sigma bonds. The concept of sigma bonding is expanded to include bonding interactions where a single lobe of one orbital overlaps with a single lobe of another. Propane, for example, is made up of ten sigma bonds, one for each of the two CC bonds and one for each of the eight CH bonds.

Hence , the answer is true .

6 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

[Ca2+] within the endoplasmic reticulum (ER) is 10-3M. Cytosolic [Ca2+] is 10-7M. Lysosomes in mammalian cells have an internal

Pani-rosa [81]

Answer:

B. Ca2+ import into the ER because it has the steeper concentration gradient

Explanation:

ΔGt = RT㏑(C₂/C₁)

where ΔGt is the free energy change for transport; R = 8.315 J/mol; T = 298 K; C₂/C₁ is ratio of concentrations inside and outside each organelle.

For Ca²⁺ import

ΔGt = 8.315 J/mol * 298 K * ㏑(10⁻³/10⁻⁷)

ΔGt= 3.42 kJ/mol

For H⁺ import

ΔGt = 8.315 J/mol * 298 K * ㏑ (10⁻⁴/10⁻⁷)

ΔGt = 2.73 kJ/mol

From the above values, ΔGt is greater for Ca²⁺ import because it has a steeper concentration gradient

8 0

3 years ago