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2 years ago

Which type of chemical reaction does this equation represent?

Chemistry

2 answers:

sweet-ann [11.9K]2 years ago

7 0

Which equation? is it <span>2HCl(aq)+Cr(s)→H₂(g)+CrCl₂(aq)</span>

horsena [70]2 years ago

3 0

Answer:

the answer is c

Explanation:

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The light-independent reactions of photosynthesis need

Oxana [17]

Answer:

I don’t know

Explanation:

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3 years ago

Which pair of elements is in the same period? (Check ALL that apply) *

stira [4]

Answer:

D and R are in the same period

Explanation:

remember groups are vertical and periods are horizontal!

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2 years ago

Given the partial equation: NO3− Pb2 → NO2 Pb4 , balance the reaction in acidic solution using the half-reaction method and fill

Fed [463]

Answer : The balanced reaction in acidic solution is,

$2NO_3^-+1Pb^{2+}+4H^+\rightarrow 2NO_2+1Pb^{4+}+2H_2O$

Explanation :

The given partial equation is,

$NO_3^-+Pb^{2+}\rightarrow NO_2+Pb^{4+}$

First we have to separate into half reaction. The two half reactions are:

$NO_3^-\rightarrow NO_2$

$Pb^{2+}\rightarrow Pb^{4+}$

Now we have to balance the half reactions in acidic medium, we get:

$NO_3^-+2H^++1e^-\rightarrow NO_2+H_2O$ ............(1)

$Pb^{2+}\rightarrow Pb^{4+}+2e^-$ ............(2)

Now we have to balance the electrons of the half reactions. When we are multiplying the equation (1) by 2, we get

$2NO_3^-+4H^++2e^-\rightarrow 2NO_2+2H_2O$ ...........(3)

Now we have to add both the half reactions (2) and (3), we get the final balanced chemical reaction.

$2NO_3^-+1Pb^{2+}+4H^+\rightarrow 2NO_2+1Pb^{4+}+2H_2O$

3 0

2 years ago

How does water evaporating from a glass show that matter is made up of particles?

UkoKoshka [18]

I think that the answer is probably b-the liquid water particles turn into water vapor that spreads in the air

6 0

3 years ago

Read 2 more answers

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