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What is the maximum amount of water (in grams) that can be removed from 15ml of toluene by the addition?

Nataly [62]

Complete Question

Magnesium sulfate forms a hydrate with the formula $MgSO_4. 7H_20$ . What is the maximum amount of water (in grams) that can be removed from 15 ml of toluene by the addition of 200 mg of anhydrous magnesium sulfate? The molar mass of $MgSO_4$ is 120.4 g/mol; H20 = 18 g/mol.

Answer:

The value is $z = 0.2093 \ g$ of $H_2O$

Explanation:

From the question we are told that

The volume of toluene is $V = 15 mL$

The mass of anhydrous magnesium sulfate is $m = 200m g = 200 *10^{-3} \ g$

The formula of the hydrate is $MgSO_4. 7H_20$

The molar mass of $MgSO_4$ is $z =120.4 \ g/mol$

From the formula given we see that

1 mole of $Mg SO_4$ wil remove 7 moles of $H_2O$ to for the given formula

Hence

120.4 g (1 mole) will remove 7 moles (7 * 18 g = 126 g ) of $H_2O$ to for the given formula

Therefore 1 g of $Mg SO_4$ x g of $H_2O$

So

$x = \frac{x]126 * 1}{ 120.4 }$

=> $x = 1.0465 \ g$

From our calculation we obtained that

1 g of $Mg SO_4$ will remove $x = 1.0465 \ g$ of $H_2O$

Then

$200 *10^{-3} \ g$ of $Mg SO_4$ will remove z g of $x = 1.0465 \ g$ of $H_2O$

So

$z = 200 *10^{-3} * 1.0465$

=> $z = 200 *10^{-3} * 1.0465$

=> $z = 0.2093 \ g$ of $H_2O$

6 0

3 years ago

What happens to wavelength as wave frequency increases?

KengaRu [80]

Answer:

As a wavelength increases in size, its frequency and energy (E) decrease. From these equations you may realize that as the frequency increases, the wavelength gets shorter. As the frequency decreases, the wavelength gets longer.

Explanation:

6 0

3 years ago

If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of s

AnnyKZ [126]

Answer:

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

$Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)$

Moles of mercury(II) acetate = $\frac{45.101 g}{318.70 g/mol}=0.14152 mol$

Moles of sodium dichromate = $\frac{12.026 g}{261.97 g/mol}=0.045906 mol$

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.

3 0

4 years ago

If He gas has an average kinetic energy of 4310 J/mol under certain conditions, what is the root mean square speed of O2 gas mol

Free_Kalibri [48]

Answer:

The root mean square speed of O2 gas molecules is

519.01 m/s

Explanation:

The root mean square velocity :

$v_{rms}=\sqrt{\frac{3RT}{M}}$

$K.E_{avg}=\frac{3}{2}RT$

$K.E =\frac{1}{2}mv_{rms}^{2}$

Molar mass , M

For He = 4 g/mol

For O2 = 2 x 16 = 32 g/mol

O2 = 32/1000 = 0.032 Kg/mol

First calculate the temperature at which the K.E of He is 4310J/mol

K.E of He =

$K.E_{avg}=\frac{3}{2}RT$

$T=\frac{2(K.E)}{3(R)}$

K.E of He = 4310 J/mol

$T=\frac{2(4310J/mol)}{3(8.314J/Kmol)}$

$T=345.60K$

Now , Use Vrms to calculate the velocity of O2

$v_{rms}=\sqrt{\frac{3(8.314J/Kmol)(345.60K)}{0.032Kg/mol}}$

$v_{rms}=\sqrt{\frac{8619.9552}{0.032}}$

$v_{rms}=\sqrt{26935.001}$

$v_{rms}=519.01m/s$

6 0

3 years ago

What are the coordination numbers of li+ and s2− in the lithium sulfide crystal shown? enter the values in order of li+ and s2−,

Anna35 [415]

Answer:
As there must be twice as many lithium ions as sulphide ions, the 'numbers'
must be different.
It's 4:8. i.e. 4 sulphides around each lithium; and 8 lithiums around each
sulphur.
It's an antifluorite structure.

5 0

3 years ago

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