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2 years ago

If you were presented with 2l of a 2m sucrose stock solution, how many grams of sugar would be in a 100 ml aliquot?

1 answer:

4 0

The 2 L of sucrose stock solution would contain similar concentration with the 100 mL aliquot. Therefore the concentration of aliquot is still 2 M.

The molar mass of sucrose is 342.3 g / mol. Therefore the mass in a 100 mL (0.1 L) aliquot is:

mass = (2 mol / L) * 0.1 L * (342.3 g / mol)

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How many moles of copper Il sulfide can be produced from 2.8 moles of sulfur? Cu + S --> CuS

Kamila [148]

Answer: 2.8 moles of copper (Il) sulfide $(CuS)$ will be produced from 2.8 moles of sulphur.

Explanation:

The balanced chemical reaction is:

$Cu+S\rightarrow CuS$

According to stoichiometry :

1 mole of $S$ produce = 1 mole of copper (Il) sulfide $(CuS)$

Thus 2.8 moles of $S$ will produce= $\frac{1}{1}\times 2.8=2.8moles$ of copper (Il) sulfide $(CuS)$

Thus 2.8 moles of copper (Il) sulfide $(CuS)$ will be produced from 2.8 moles of sulphur.

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2 years ago

You have been given an orange liquid and told to describe its properties.

chubhunter [2.5K]

Answer:

The correct answer would be - observing with the help of five senses.

Explanation:

To find and describe the physical properties of the given substance or the solution or liquid students can observe using their five senses. By looking at the liquid one can find its state and color, by smelling students can find the odor of the sample, by touching it one can observe and describe the texture.

Fluidity can also be measure by the touch if the solution is viscous or free-floating. By using a thermometer and using a graduated cylinder one can find the temperature at room temperature and the weight of substance respectively.

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3 years ago

Which statement best describes scientific laws? Scientific laws are facts that describe natural events. Scientific laws are theo

Fiesta28 [93]

Well scientific laws are really laws! You can't change a scientific law and laws can be facts as well. Hope this helps!

7 0

2 years ago

Read 2 more answers

Click the button that shows the correct relationship of the electron affinities of the elements sodium and phosphorus.

VikaD [51]

Answer:

The second option

Explanation:

I took the quiz and got it correct

3 0

2 years ago

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Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

3 years ago