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3 years ago

Harry Hess resurrected Wegener’s continental drift hypothesis and also the mantle convection idea of Holmes.

Chemistry

2 answers:

stepan [7]3 years ago

7 0

Answer:

true

Explanation:

aleksklad [387]3 years ago

3 0

He is right its True

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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

(b) Ozone absorbs radiation of wavelengths 2200–2900 Å, thus protecting organisms from this radiation. Find the frequency and en

GrogVix [38]

The frequency and energy of most energetic of these photons is $1.36*10^{15}$ Hz and $9.04*10^{-19}$ J respectively.

The high energy photons will have least wavelength. Taking the given less wavelength for calculation of frequency and energy of photons.

λ = 2200 Å

Also, we know, 1 Å = $10^{-10}$ m

So, λ = $2200*10^{-10}$

λ = $2.2*10^{-7}$ <em> </em>m

Now, calculating frequency using the formula -

c = fλ, where c is speed of light, f is frequency and λ is wavelength.

We know, c = $3*10^{8}$ m/s

f = c/λ

f = $\frac{3*10^{8} }{2.2*10^{-7} }$

frequency = $1.36*10^{15}$ Hz

Now, calculating Energy through the formula -

E = hf, where E is energy, h is planck's constant and f is frequency

Value of planck's constant = $6.63*10^{-34}$ Js

E = $6.63*10^{-34}$ * $1.36*10^{15}$

Energy = $9.04*10^{-19}$ J

Thus, frequency and energy of photons is $1.36*10^{15}$ Hz and $9.04*10^{-19}$ J respectively.

Learn more about energy of photon -

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3 0

1 year ago

Match the terms with their correct definition:

geniusboy [140]

7 0

3 years ago

A sample consisting of n mol of an ideal gas undergoes a reversible isobaric expansion from volume Vi to volume 3Vi. Find the ch

kolezko [41]

Answer:

The change in entropy of gas is $\Delta S= nC_{P}ln3$

Explanation:

n= Number of moles of gas

Change in entropy of gas = $ds= \int \frac{dQ}{T}$

$dQ= nC_{p}dT$

From the given,

$V_{i}=V$

$V_{f}=3V$

Let "T" be the initial temperature.

$\frac {V_{i}}{T_{i}}=\frac {V_{f}}{T_{f}}$

$\frac {V}{T}=\frac {3V}{T_{f}}$

${T_{f}} = 3T$

$\int ds = \int ^{T_{f}}_{T_{i}} \frac{nC_{P}dT}{T}$

$\Delta S = nC_{p}ln(\frac{T_{f}}{T_{i}})$

$\Delta S = nC_{p}ln3$

Therefore, The change in entropy of gas is $\Delta S= nC_{P}ln3$

3 0

3 years ago

For the following reaction, 5.22 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 12.9 grams of

Mamont248 [21]

Answer:

$m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3$

Explanation:

Hello,

In this case, the undergoing balanced chemical reaction is:

$Al_2O_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2O(l)$

Thus, as 5.22 grams of aluminium oxide reacts, the required yielded amount of aluminium sulfate results:

$m_{Al_2(SO_4)_3}=5.22gAl_2O_3*\frac{1molAl_2O_3}{102gAl_2O_3}*\frac{1molAl_2(SO_4)_3}{1molAl_2O_3}*\frac{342gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3$

Moreover, the percent yield is:

$Y=\frac{12.9g}{17.5g} *100\%=73.7\%$

Best regards.

6 0

4 years ago