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3 years ago

URGENT PLEASE HELP!!!

2 answers:

mote1985 [20]3 years ago

5 0

Even though two grams seemed to disappear or vanish, the law of conversation of mass still holding true. Mercuric oxide, when heated, forms a gas of mercury and oxygen. During the investigation, some gas could have escaped or evaporated.

Annette [7]3 years ago

5 0

Answer:Yes, the law of conservation of mass is true because even though some/2grames of mass was not there to end the 2 grams might have evaporated or escaped.

Explanation:

I got this right on flvs and more explanation is up there kinda. hope this helps

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In which state of matter are water molecules measured as having a comparatively high temperature?

skelet666 [1.2K]

Answer:

water vapor

because solid ice is a matter

5 0

3 years ago

Question 2 of 25

DedPeter [7]

Answer:

The right objective is A I guess

8 0

3 years ago

A sample of gaseous methyl ether has a mass of 8.12g and occupies a volume of 3.96L at STP. What is the molecular weight of meth

Naya [18.7K]

Molar volume is when 1 mol of any gas occupies 22.4 L at STP.
Methyl ether has a mass of 8.12 g,
Volume occupied - 3.96 L
If 22.4 L occupied by 1 mol of gas
Then 3.96 L occupied by 1/22.4 x 3.96 = 0.176 mol of gas
The mass of 0.176 mol = 8.12 g
Molar mass is mass of 1 mol
Therefore mass of 1 mol = 8.12/0.176 = 46.1
Molecular weight is 46.1 g/mol

5 0

3 years ago

You add 0.5 moles of NaCl to water to make a 250 mL solution. What is the molarity

evablogger [386]

Answer:

The correct answer is 2, 0M

Explanation:

We calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter):

250 ml solution----- 0,5 moles of NaCl

1000 ml solution----x= (1000 ml solution x 0,5 moles of NaCl)/250 ml solution

x= 2,0 moles of NaCl --> <em>The solution is 2 M</em>

5 0

3 years ago

Calculate the frequency of the n=2 line in the lyman series of hydrogen

Alona [7]

Answer:

Approximately $2.47\times 10^{15}\; \rm Hz$ .

Explanation:

The Lyman Series of a hydrogen atom are due to electron transitions from energy levels $n \ge 2$ to the ground state where $n = 1$ . In this case, the electron responsible for the line started at $n = 2$ and transitioned to

A hydrogen atom contains only one electron. As a result, Bohr Model provides a good estimate of that electron's energy at different levels.

In Bohr's Model, the equation for an electron at energy level $n$ (

$\displaystyle - \frac{k\, Z^2}{n^2}$ (note the negative sign in front of the fraction,)

where

$k = 2.179 \times 10^{-18}\; \rm J$ is a constant.

$Z$ is the atomic number of that atom. $Z = 1$ for hydrogen.
$n$ is the energy level of that electron.

The electron that produced the $n = 2$ line was initially at the

$\begin{aligned} &E_{n = 2} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{2^2} \cr & \approx -5.4475\times 10^{-19}\; \rm J\end{aligned}$ .

The electron would then transit to energy level $n = 1$ . Its energy would become:

$\begin{aligned} &E_{n = 1} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{1^2} \cr & \approx -2.179 \times 10^{-18} \; \rm J\end{aligned}$ .

The energy change would be equal to

$\begin{aligned}&\text{Initial Energy} - \text{Final Energy} \cr &= E_{n = 2} - E_{n = 1} \cr &= -5.4475 \times 10^{-19} - \left(-2.179 \times 10^{-18}\right) \cr & \approx 1.63425\times 10^{-18}\; \rm J \end{aligned}$ .

That would be the energy of a photon in that $n = 2$ spectrum line. Planck constant $h$ relates the frequency of a photon to its energy:

$E = h \cdot f$ , where

$E$ is the energy of the photon.
$h \approx 6.62607015\times 10^{-34}\; \rm J \cdot s$ is the Planck constant.
$f$ is the frequency of that photon.

In this case, $E \approx 1.63425 \times 10^{-18}\; \rm J$ . Hence,

$\begin{aligned} f &= \frac{E}{h} \cr &\approx \frac{1.63425\times 10^{-18}}{6.62607015\times 10^{-34}} \cr & \approx 2.47 \times 10^{15}\; \rm s^{-1}\end{aligned}$ .

Note that $1 \; \rm Hz = 1 \; \rm s^{-1}$ .

6 0

3 years ago