Compare the the volumes and densities of two pieces of lead: one with a mass of 25 g and the other with a mass of 75 g.

A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r

Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

$P\cdot V = n\cdot R\cdot T$ ,

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of gas particles in the gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume $V$ of the gas. Rearrange the ideal gas equation for volume:

$V = \dfrac{n \cdot R \cdot T}{P}$ .

Both the temperature of the gas, $T$ , and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

$T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K}$ ,
$T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}$ .

The volume of the gas is proportional to its temperature if both $n$ and $P$ stay constant.

$n$ won't change unless the balloon leaks, and
consider $P$ to be constant, for calculations that include $T$ .

$V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}$ .

Now, keep the temperature at $T_1 =258.65\;\text{K}$ and change the pressure on the gas:

$P_1 = 0.995\;\text{atm}$ ,
$P_2 = 0.720\;\text{atm}$ .

The volume of the gas is proportional to the reciprocal of its absolute temperature $\dfrac{1}{T}$ if both $n$ and $T$ stays constant. In other words,

$V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}$

(3 sig. fig. as in the question.).

See if you get the same result if you hold $T$ constant, change $P$ , and then move on to change $T$ .

6 0

3 years ago

Why is it important to collect all data during experiments, rather than make inferences of what will happen?

Paraphin [41]

Answer:

It is important to collect all data first, or else your guesses could purely be the opposite of the right answer. If you make inferences of what might happen, your guesses may be purely fictional, and totally off-topic. During experiments, this step is important.

7 0

3 years ago

1. An acid will have the suffix “–ic acid” at the end of its name when the negative

Nimfa-mama [501]

An acid will have the suffix “–ic acid” at the end of its name when the negative ion has a suffix -ide.

An acid will have the suffix “–ous acid” at the end of its name when the negative ion has a suffix -ite.

This is defined as a chemical substance which can react with a base to form a salt in a reaction.

The appropriate acid suffixes and conditions is however correctly given above.

Read more about Acid here brainly.com/question/2183107

4 0

2 years ago

Group 2A metal oxides react with water to produce hydroxides. When 35.55 g of CaO are placed into 125 mL of water (d= 1.000 g/mL

Vika [28.1K]

Answer is: mass of calcium hydroxide is 46.98 grams.
Balanced chemical reaction: CaO + H₂O → Ca(OH)₂.
m(CaO) = 35.55 g.
n(CaO) = m(CaO) ÷ M(CaO).
n(CaO) = 35.55 g ÷ 56 g/mol.
n(CaO) = 0.634 mol; limiting reactant.
m(H₂O) = 125 mL · 1.000 g/mL.
m(H₂O) = 125 g.
n(H₂O) = 125 g ÷ 18 g/mol.
n(H₂O) = 6.94 mol.
From chemical reaction: n(CaO) : n(Ca(OH)₂) = 1 : 1.
n(Ca(OH)₂) = 0.634 mol.
m(Ca(OH)₂) = 0.634 mol · 74.1 g/mol = 46.98 g.

5 0

3 years ago

Describe the results of the experiment that proved atoms contain positive,negative, and neutral particles

Bezzdna [24]

Explanation:

J.J. Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons. ... Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus.

there u go :>

5 0

4 years ago