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2 years ago

Which statement is true about the particles of a solid when the solid

Chemistry

2 answers:

Vikki [24]2 years ago

8 0

2) Their kinetic energy increase.

ira [324]2 years ago

6 0

Answer:

2) Their kinetic energy increases.

Explanation:

The particles have more kinetic energy because they move around more when they are a liquid.

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HELP NOW BRAINLIST AND 15 POINTS!!!!!! PLEASE HELPPP

ycow [4]

Answer:

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2 years ago

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One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c

Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

$n(\mathrm{O_2}) = n(\mathrm{H}_2)$ .

Note that the mass of a gas $m$ is different from the number of gas particles $n$ in it. In particular, if all particles in this gas have a molar mass of $M$ , then:

$m = n \cdot M$ .

In other words,

$m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2})$ .
$m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2})$ .

The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

Look up relative atomic mass data on a modern periodic table:

$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Therefore:

$M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}$ .
$M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}$ .

Verify whether $\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ :

Left-hand side: $\displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1$ .
Right-hand side: $\displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9$ .

Note that the mass of the $\rm H_2\; (g)$ sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

7 0

3 years ago

What mass of water will change its temperature by 3.0 0C when 525 J of heat is added to it? The specific heat of water is 4.184

pav-90 [236]

Answer:

Mass of water = 41.8 g

Explanation:

Given data:

Mass of water = ?

Change in temperature = 3.0 °C

Specific heat capacity = 4.184 j/g.°C

Heat absorbed = 525 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 3.0°C

Now we will put the values in formula.

525 J = m × 4.184 j/g.°C × 3.0°C

525 J = m × 12.552 j/g

m = 525 J/ 12.552 j/g

m = 41.8 g

4 0

3 years ago

put the contributions to the understanding of the atomic structure in order from most recent at the top to the earliest at the b

r-ruslan [8.4K]

Answer:

From Top to Bottom:

- Democritus coming up with the concept of an atom

- Dalton discovering that atoms are the smallest part of an element

- Rutherford discovering the nucleus of an atom

- Thomson discovering electrons

- Bohr modeling electrons orbiting the nucleus

- Schrodinger modeling electrons in the electron cloud

Explanation:

The best way to think about this is from the inside out. Democrats (who lived long before any of the other scientists mentioned) was the one who thought of the idea of the atom. - Therefore, this must be first because all other choices are elaborations on the idea that atoms exist. Next must be Dalton. Dalton saw atoms as "cannonballs" if you will; a solid mass. So then after that, Rutherford and his gold foil experiment (he discovered that some rays he shot through gold foil were deflected back; ie the existence of concentrated areas in an atom, ie the nucleus). Then we get into the information on electrons. We must start with discovery (Thomson). Heres where it gets complicated. Electrons don't <em>actually </em>orbit the nucleus, they exist in electron clouds. So it would be Bohr, who came up with the idea that electron exist outside the nucleus, then Schrodinger, who elaborated on Bohr's theory. Hope this helps!

Nat, Junior

Accel + AP Chem student

5 0

1 year ago

The substances that are produced after the chemical reaction are called the ___

sertanlavr [38]

Answer:

Products

Explanation:

During a chemical reaction, there are the reactants (left side), and the products (right side).

6 0

2 years ago

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