All categories

3 years ago

PLZ help me

Chemistry

1 answer:

Alona [7]3 years ago

5 0

Answer:

Wavelength in high frequency waves:

Wavelength. Waves of blue light have a shorter wavelength than waves of red light. The frequency of a wave is inversely proportional to its wavelength. That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength.

Wavelength in low frequency waves:

The frequency of a wave is inversely proportional to its wavelength. That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength. Light waves have very, very short wavelengths.

Explanation:

You might be interested in

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

3 years ago

Plz answer 3 ,4,5 thanks

ratelena [41]

For #3 the answer is B, for #4 the answer is F, and for #5 the answer is B.

5 0

3 years ago

The half life of Radon-222 is 3.8 days. If you have a 90.0g sample of Radon-222 in the lab, how much will be left after 19 days?

statuscvo [17]

Answer:

half lives passed=5

given sample=90g

sample left=2.8125g

Explanation:

no. of half lives=total time/half life

no.=19days/3.8days

no.=5 days

after 5 half lives sample left=2.8125g

5 0

3 years ago

Which of the following ions would represent the ion of an element from Group 2A?

Gennadij [26K]

Answer:

D. A²⁺

Explanation:

Element from Group 2A -

The element of group 2A are known as the alkaline earth metals ,

The elements of this group are - beryllium (Be) , magnesium (Mg) , calcium (Ca) , strontium (Sr) , barium (Ba) , and radium (Ra).

The valence shell of these elements have two valence electrons and therefore , if the element need to stabilizes itself , it need to lose these two electrons and attain noble gas configuration,

Therefore ,

After losing the two electrons , the element becomes A²⁺.

Hence , the ions are symbolized as A²⁺.

3 0

3 years ago

Choose all the answers that apply.

mina [271]

Is created by the movement of electrons

6 0

4 years ago