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nirvana33 [79]
3 years ago
10

HELP I think its C but I'm not sure

Chemistry
1 answer:
nasty-shy [4]3 years ago
4 0
The answer is A. Mars

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Why does increasing concentration usually increase reaction rate?
Alex Ar [27]
The correct answer is C) There are more particle collision

With more particle collision, more reactions are created.
3 0
3 years ago
Read 2 more answers
The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen
tino4ka555 [31]

Answer: The mass percentage of Tl_2SO_4 is 5.86%

Explanation:

To calculate the mass percentage of Tl_2SO_4 in the sample it is necessary to know the mass of the solute (Tl_2SO_4 in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the TlI precipitate.  We can establish a relation between the mass of TlI and Tl_2SO_4 using the stoichiometry of the compounds:

moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.

Since for every mole of Tl in TlI there are two moles of Tl in Tl_2SO_4, we have:

moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol

Using the molar mass of Tl_2SO_4 we have:

mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g

Finally, we can use the mass percentage formula:

mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%

6 0
3 years ago
What is my control group in my experiment and why is it that?
alex41 [277]
Your control group would be the batteries since you CONTROL what brand you're using, for which one lasts the longest...aren't you suppose to figure that out when you do the experiment?
8 0
3 years ago
Read 2 more answers
Calculate the volume of a gas at a pressure of 150 kpa if its volume at 180 kpa is 1500 L
Anastaziya [24]

Answer: 1800 L

Explanation:

Given that,

Original pressure of gas (P1) = 180 kPa

Original volume of gas (V1) = 1500 L

New pressure of gas (P2) = 150 kPa

New volume of gas (V2) = ?

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

P1V1 = P1V2

180 kPa x 1500 L = 150 kPa x V2

270000 kPa•L = 150 kPa•V2

Divide both sides by 150 kPa

270000 kPa•L/150 kPa = 150 kPa•V2/150 kPa

1800L = V2

Thus, the new volume of the gas is 1800 liters.

8 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
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